This chapter delves into the fundamental structure of atoms, the building blocks of matter. It covers the following key points:

**Subatomic Particles**

**Discovery of electrons:**J.J. Thomson’s experiments led to the discovery of electrons, negatively charged particles.**Discovery of protons:**Ernest Rutherford’s gold foil experiment revealed the existence of a positively charged nucleus.**Neutrons:**James Chadwick discovered neutrons, neutral particles in the nucleus.**Properties of subatomic particles:**Mass and charge of protons, neutrons, and electrons.

**Atomic Models**

**Thomson’s model:**Proposed a plum pudding model with a positive sphere and embedded electrons.**Rutherford’s model:**Suggested a nuclear model with a small, dense, positively charged nucleus and electrons orbiting around it.**Bohr’s model:**Introduced quantized energy levels for electrons, explaining the stability of atoms.

**Atomic Number and Mass Number**

**Atomic number (Z):**The number of protons in the nucleus, determining the identity of an element.

**Electronic Configuration**

**Quantum numbers:**Four sets of numbers (n, l, m, and s) that describe the state of an electron.**Hund’s rule:**Orbitals within a subshell are filled singly before pairing.

**In essence, this chapter provides a comprehensive understanding of the structure of atoms, including the subatomic particles, atomic models, and the arrangement of electrons in orbitals.** It also introduces key concepts like atomic number, mass number, and electronic configuration.

**Exercise **

**1. ****(i) Calculate the number of electrons which will together weigh one gram.**

**(ii) Calculate the mass and charge of one mole of electrons.**

**Ans : **To calculate the number of electrons that weigh one gram, we need to know the mass of a single electron.

Given:

Mass of one electron = 9.1096 × 10⁻³¹ kg

Conversion:

1 kg = 10³ g

Mass of one electron = 9.1096 × 10⁻³¹ kg * 10³ g/kg = 9.1096 × 10⁻²⁸ g

Calculation:

Let the number of electrons needed to weigh one gram be x.

x * 9.1096 × 10⁻²⁸ g = 1 g

x = 1 g / (9.1096 × 10⁻²⁸ g)

**x ≈ 1.098 × 10²⁷ electrons **

**Therefore, approximately 1.098 × 10²⁷ electrons would weigh one gram**

**(ii) 1. Mass:**

Mass of one electron: 9.1096 × 10⁻³¹ kg

Moles of electrons in one mole: 6.022 × 10²³ electrons/mol (Avogadro’s number)

Mass of one mole of electrons: 9.1096 × 10⁻³¹ kg/electron * 6.022 × 10²³ **electrons/mol ≈ 5.486 × 10⁻⁷ kg/mol**

**2. Charge:**

Charge of one electron: -1.602 × 10⁻¹⁹ C

Charge of one mole of electrons: -1.602 × 10⁻¹⁹ C/electron * 6.022 × 10²³ **electrons/mol ≈ -9.648 × 10⁴ C/mol**

**Therefore:**

Mass of one mole of electrons: 5.486 × 10⁻⁷ kg/mol

**Charge of one mole of electrons =**

** -9.648 × 10⁴ C/mol**

**2.** **(i) Calculate the total number of electrons present in one mole of methane.**

**(ii) Find (a) the total number and**

**(b) the total mass of neutrons in 7 mg of 14C.**

** (Assume that mass of a neutron = 1.675 × 10-27kg).**

**(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.**

**Will the answer change if the temperature and pressure are changed ?**

**Ans : **

**(i) Total number of electrons in one mole of methane (CH₄)**

**Electrons in one CH₄ molecule:**- Carbon (C): 6 electrons
- Hydrogen (H): 1 electron each (4 hydrogen atoms)
- Total electrons in one CH₄ molecule: 6 + 4 = 10 electrons

**Electrons in one mole of CH₄:**- 1 mole contains 6.022 × 10²³ molecules
- Total electrons = 10 electrons/molecule * 6.022 × 10²³ molecules/mole = 6.022 × 10²⁴ electrons

**(ii) Total number and mass of neutrons in 7 mg of ¹⁴C**

**Number of neutrons in one ¹⁴C atom:**- Mass number (A) = 14
- Atomic number (Z) = 6 (number of protons)
- Number of neutrons = A – Z = 14 – 6 = 8

**Moles of ¹⁴C in 7 mg:**- Molar mass of ¹⁴C ≈ 14 g/mol
- Moles = mass / molar mass = 7 mg / 14 g/mol = 0.0005 moles

**Total number of neutrons:**- 0.0005 moles * 8 neutrons/atom * 6.022 × 10²³ atoms/mole ≈ 2.409 × 10²¹ neutrons

**Total mass of neutrons:**- Mass of one neutron = 1.675 × 10⁻²⁷ kg
- Total mass = 2.409 × 10²¹ neutrons * 1.675 × 10⁻²⁷ kg/neutron ≈ 4.03 × 10⁻⁶ kg

**(iii) Total number and mass of protons in 34 mg of NH₃ at STP**

**Protons in one NH₃ molecule:**- Nitrogen (N): 7 protons
- Hydrogen (H): 1 proton each (3 hydrogen atoms)
- Total protons = 7 + 3 = 10 protons

**Moles of NH₃:**- Molar mass of NH₃ ≈ 17 g/mol
- Moles = mass / molar mass = 34 mg / 17 g/mol = 0.002 moles

**Total number of protons:**- 0.002 moles * 10 protons/molecule * 6.022 × 10²³ molecules/mole ≈ 1.204 × 10²² protons

**Total mass of protons:**- Mass of one proton ≈ 1.673 × 10⁻²⁷ kg
- Total mass ≈ 1.204 × 10²² protons * 1.673 × 10⁻²⁷ kg/proton ≈ 2.01 × 10⁻⁵ kg

**3. How many protons and neutrons are present in the following nuclei**

**Ans : **

**¹³₆C**

- Atomic number (Z) = 6
- Mass number (A) = 13
- Number of protons = 6
- Number of neutrons = 13 – 6 = 7

**¹⁶₈O**

- Atomic number (Z) = 8
- Mass number (A) = 16
- Number of protons = 8
- Number of neutrons = 16 – 8 = 8

**²⁴₁₂Mg**

- Atomic number (Z) = 12
- Mass number (A) = 24
- Number of protons = 12
- Number of neutrons = 24 – 12 = 12

**⁵⁶₂₆Fe**

- Atomic number (Z) = 26
- Mass number (A) = 56
- Number of protons = 26
- Number of neutrons = 56 – 26 = 30

**⁸⁸₃₈Sr**

- Atomic number (Z) = 38
- Mass number (A) = 88
- Number of protons = 38
- Number of neutrons = 88 – 38 = 50

**4. Write the complete symbol for the atom (X) with the given atomic number (Z) and atomic mass (A)**

**(i) Z = 17,A = 35**

**(ii) Z = 92, A = 233**

**(in) Z = 4, A = 9.**

**Ans : **

**(i) ³⁵₁₇Cl**

**(ii)** **²³³₉₂U**

**(iii)** **⁹₄Be**

**5. Yellow light emitted from a sodium lamp has a wavelength (2) of 580 nm. Calculate the frequency (v) and wave number (v) of yellow light.**

**Ans : **

**Given:**

- Wavelength (λ) = 580 nm = 580 × 10⁻⁹ m

**Constants:**

- Speed of light (c)
- = 3 × 10⁸ m/s

**Formula:**

- Frequency (ν) = c / λ
- Wavenumber (ν̄) = 1 / λ

**Calculations:**

**Frequency (ν):**- ν = (3 × 10⁸ m/s) / (580 × 10⁻⁹ m)
- ν ≈ 5.17 × 10¹⁴ s⁻¹

**Wavenumber (ν̄):**- ν̄ = 1 / (580 × 10⁻⁹ m)
- ν̄ ≈ 1.72 × 10⁶ m⁻¹

**Therefore, the frequency of the yellow light is approximately 5.17 × 10¹⁴ s⁻¹ and the wavenumber is approximately 1.72 × 10⁶ m⁻¹.**

**6. Calculate the energy of each of the photons which**

**(i) correspond to light of frequency 3 × 1015 Hz**

**(ii) have wavelength of 0-50 A.**

**Ans : **

E = hν

where:

- E is the energy of the photon (in joules)
- h is Planck’s constant (6.626 × 10⁻³⁴ J s)
- ν is the frequency of the light (in hertz)

Alternatively, we can use the following equation if we know the wavelength (λ) of the light:

E = hc / λ

**Case (i): Frequency = 3 × 10¹⁵ Hz**

E = (6.626 × 10⁻³⁴ J s) × (3 × 10¹⁵ Hz)

E ≈ 1.99 × 10⁻¹⁸ J

**Case (ii): Wavelength = 0.50 Å = 0.50 × 10⁻¹⁰ m**

E = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (0.50 × 10⁻¹⁰ m)

**E ≈ 3.98 × 10⁻¹⁵ J**

**7. Calculate the wavelength, frequency, and wavenumber of lightwave whose period is 2.0 × 10****-10**** s.**

**Ans : **

**1. Calculate the frequency (ν):**

Frequency is the reciprocal of the period (T).

ν = 1 / (2.0 × 10⁻¹⁰ s)

**ν = 5.0 × 10⁹ s⁻¹**

**2. Calculate the wavelength (λ):**

c = νλ

Therefore, we can rearrange this equation to solve for λ:

λ = c / ν

λ = (3.0 × 10⁸ m/s) / (5.0 × 10⁹ s⁻¹)

**λ = 6.0 × 10⁻² m**

**3. Calculate the wavenumber (ν̄):**

The wavenumber is the reciprocal of the wavelength.

ν̄ = 1 / λ

ν̄ = 1 / (6.0 × 10⁻² m)

**ν̄ = 16.7 m⁻¹ **

**8.What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?**

**Ans : **

To calculate the number of photons, we’ll first need to find the energy of a single photon.

**Formula for energy of a photon:**

**E = hc / λ**

where:

- E = energy of the photon (in joules)
- h = Planck’s constant (6.626 × 10⁻³⁴ J s)
- c = speed of light (3.00 × 10⁸ m/s)
- λ = wavelength of the light (in meters)

**1. Converting wavelength to meters:**

4000 pm = 4000 × 10⁻¹² m = 4.0 × 10⁻⁹ m

**2.Calculating energy of a single photon:**

E = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (4.0 × 10⁻⁹ m) E ≈ 4.97 × 10⁻¹⁷ J

**3.Calculating the number of photons:**

**Number of photons = Total energy / Energy per photon**

Number of photons = 1 J / (4.97 × 10⁻¹⁷ J/photon) Number of photons ≈ 2.01 × 10¹⁶ photons

**Therefore, approximately 2.01 × 10¹⁶ photons of light with a wavelength of 4000 pm are required to provide 1 J of energy.**

**9.A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J)**

**Ans :** Understanding the Problem

We’re given the wavelength of a photon and the work function of a metal. We need to calculate:

The energy of the photon (in eV)

The kinetic energy of the emitted photoelectron

The velocity of the photoelectron

Formula:

**1.Energy of a photon (E): E = hc / λ **

**Kinetic energy of the photoelectron (KE): KE = E – Φ**

**Velocity of the photoelectron (v): KE = (1/2)mv², where m is the mass of an electron**

Constants:

(h) Planck’s constant = 6.626 × 10⁻³⁴ J s

(c) Speed of Light = 3.00 × 10⁸ m/s

Mass of an electron (m) = 9.11 × 10⁻³¹ kg

Calculations:

Energy of the photon (E):

E = (6.626 × 10⁻³⁴ J s) * (3.00 × 10⁸ m/s) / (4 × 10⁻⁷ m)

E ≈ 4.97 × 10⁻¹⁹ J

**Since 1 eV = 1.6020 × 10⁻¹⁹ J, E ≈ 3.10 eV**

**2.Kinetic energy of the emission (KE):**

KE = E – Φ

KE = 3.10 eV – 2.13 eV

**KE = 0.97 eV**

**3.Velocity of the photoelectron (v):**

First, convert KE to joules: 0.97 eV * 1.6020 × 10⁻¹⁹ J/eV ≈ 1.55 × 10⁻¹⁹ J

Then, use the kinetic energy formula:

(1/2)mv² = 1.55 × 10⁻¹⁹ J

v² = (2 * 1.55 × 10⁻¹⁹ J) / (9.11 × 10⁻³¹ kg)

**v ≈ 5.85 × 10⁵ m/s**

Therefore:

**(i) The energy of the photon is approximately 3.10 eV.**

**(ii) The kinetic energy of the emitted photoelectron is approximately 0.97 eV. **

**(iii) The velocity of the photoelectron is approximately 5.85 × 10⁵ m/s.**

**10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.**

**Ans : **

To calculate the ionisation energy of sodium, we first need to find the energy of a single photon with a wavelength of 242 nm. Then, we’ll convert this energy to kJ/mol.

**Formula for energy of a photon:**

**E = hc / λ**

where:

- E = energy of the photon (in joules)
- h = Planck’s constant (6.626 × 10⁻³⁴ J s)
- c = speed of light (3.00 × 10⁸ m/s)
- λ = wavelength of the light (in meters)

**Converting wavelength to meters:**

242 nm = 242 × 10⁻⁹ m

**Calculating energy of a single photon:**

E = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (242 × 10⁻⁹ m) E ≈ 8.21 × 10⁻¹⁹ J

**Converting energy to kJ/mol:**

1 mole of sodium atoms contains 6.022 × 10²³ atoms.

Ionisation energy per mole = (8.21 × 10⁻¹⁹ J/atom) × (6.022 × 10²³ atoms/mol) × (1 kJ / 1000 J)

Ionisation energy per mole ≈ 494 kJ/mol

**Therefore, the ionisation energy of sodium is approximately 494 kJ/mol.**

**11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.**

**Ans : ** To calculate the rate of emission of quanta per second, we’ll first need to find the energy of a single photon and then divide the total power of the bulb by this energy.

**Formula for energy of a photon:**

**E = hc / λ**

where:

E = energy of the photon (in joules)

h = Planck’s constant (6.626 × 10⁻³⁴ J s)

c = speed of light (3.00 × 10⁸ m/s)

λ = wavelength of the Light (in meters)

**convert the wavelength from nm to meters**

0.57 μm = 0.57 × 10⁻⁶ m

**Calculating energy of a single photon:**

E = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (0.57 × 10⁻⁶ m)

**E ≈ 3.49 × 10⁻¹⁹ J **

**Calculating the rate of emission of quanta:**

Rate of emission = Power / Energy per photon

Rate of emission = 25 W / (3.49 × 10⁻¹⁹ J/photon)

**Rate of emission ≈ 7.17 × 10¹⁹ photons/s **

**Therefore, the rate of emission of quanta per second is approximately 7.17 × 10¹⁹ photons/s**

**12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal**

**Ans : **To calculate the threshold frequency (ν₀) and work function (W₀) of the metal, we can use the following equations:

**1. Energy of a photon:**

E = hc / λ

where:

E = energy of the photon (in joules)

h = Planck’s constant (6.626 × 10⁻³⁴ J s)

c = speed of light (3.00 × 10⁸ m/s)

λ = wavelength of the light (in meters)

**2. Relationship between energy of a photon and work function:**

E = W₀ + KE

where:

W₀ = work function of the metal (in joules)

KE = kinetic energy of the emitted electron (in joules)

Since electrons are emitted with zero velocity, their kinetic energy (KE) is zero.

Here, we can simplify the equation to:

E = W₀

**1..Calculate the energy of the photon:**

λ = 6800 Å = 6800 × 10⁻¹⁰ m

E = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (6800 × 10⁻¹⁰ m)

E ≈ 2.91 × 10⁻¹⁹ J

**2.Since E = W₀, the work function (W₀) is 2.91 × 10⁻¹⁹ J.**

**3.Calculate the threshold frequency (ν₀):**

W₀ = hν₀

ν₀ = W₀ / h

ν₀ = (2.91 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J s)

ν₀ ≈ 4.40 × 10¹⁴ Hz

**Therefore, the threshold frequency (ν₀) is 4.40 × 10¹⁴ Hz and the work function (W₀) is 2.91 × 10⁻¹⁹ J.**

**13.What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2? **

**Ans : **To calculate the wavelength of light emitted when an electron in a hydrogen atom transitions from n = 4 to n = 2,

**1.we can use the Rydberg formula:**

1/λ = R_H * (1/n₁² – 1/n₂²)

where:

λ = wavelength of the emitted light

R_H is the Rydberg constant (approximately 1.097 × 10⁷ m⁻¹)

n₁ is the initial energy level (4 in this case)

n₂ is the final energy level (2 in this case)

**2.Entering the values into the formula, we get:**

1/λ =

(1.097 × 10⁷ m⁻¹) * (1/2² – 1/4²)

1/λ =

(1.097 × 10⁷ m⁻¹) * (3/16)

1/λ =

2.057 × 10⁶ m⁻¹

**3.Now, we can find λ by taking the reciprocal of both sides:**

λ = 1 / (2.057 × 10⁶ m⁻¹)

λ ≈ 4.86 × 10⁻⁷ m

**Therefore, the wavelength of the light emitted is approximately 4.86 × 10⁻⁷ meters, or 486 nanometers.**

**14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionisation enthalpy of H atom (energy required to remove the electron from n =1 orbit). **

**Ans : **To calculate the ionization energy of a hydrogen atom when the electron occupies the n = 5 orbit,

**1.we can use the following formula:**

**ΔE = -R_H * (1/n₂² – 1/n₁²)**

where:

ΔE = energy difference between the two energy levels

R_H is the Rydberg constant (approximately 2.18 × 10⁻¹⁸ J)

n₁ is the initial energy level (5 in this case)

n₂ is the final energy level (∞ for ionization)

**Entering the values into the formula,**

ΔE = -2.18 × 10⁻¹⁸ J * (1/∞² – 1/5²)

ΔE = -2.18 × 10⁻¹⁸ J * (0 – 0.04)

**ΔE ≈ 8.72 × 10⁻²⁰ J**

Since the ionization energy is the energy required to remove the electron to infinity (n = ∞), the calculated energy difference (ΔE) is the ionization energy.

**Therefore, the energy required to ionize a hydrogen atom when the electron occupies the n = 5 orbit is approximately 8.72 × 10⁻²⁰ J.**

**2.Comparison with ionization enthalpy of H atom (n = 1):**

The ionization enthalpy of a hydrogen atom (the energy required to remove the electron from the n = 1 orbit) is 2.18 × 10⁻¹⁸ J.

Comparing the two values, we can see that:

The ionization energy for n = 5 is significantly lower than the ionization enthalpy for n = 1.

This is because electrons in higher energy levels are more loosely bound to the nucleus and require less energy to remove.

In fact, the ratio of the ionization energy for n = 1 to the ionization energy for n = 5 is:

2.18 × 10⁻¹⁸ J / 8.72 × 10⁻²⁰ J ≈ **25**

**This means that the ionization energy for n = 1 is approximately 25 times greater than the ionization energy for n = 5**

**15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state? **

**Ans :**The maximum number of emission lines is calculated using the formula:

**n(n-1)/2 = 6(6-1)/2 = 3*15 = 15**

This formula indicates that the maximum number of emission lines for an element with n energy levels is equal to the product of n and (n-1) divided by 2. In this case, with n=6 energy levels, the maximum number of emission lines is 15.

**16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? **

** (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom. **

**Ans : (i) The energy associated with the nth orbit of a hydrogen atom is given by the formula:**

**En = -2.18 × 10⁻¹⁸ / n² J/atom**

where n is the principal quantum number.

For the fifth orbit, n = 5.

Entering this into the formula,

E₅ = -2.18 × 10⁻¹⁸ / 5² J/atom

E₅ = -8.72 × 10⁻²⁰ J/atom

**Therefore, the energy associated with the fifth orbit in the hydrogen atom is -8.72 × 10⁻²⁰ J/atom.**

**(ii) The radius of the nth Bohr orbit of a hydrogen atom is given by the formula:**

rn = n² * a₀

where:

a₀ is the Bohr radius (5.29 × 10⁻¹¹ m)

For the fifth orbit, n = 5.

r₅ = 5² * 5.29 × 10⁻¹¹ m

r₅ = 1.3225 × 10⁻⁹ m

**Therefore, the radius of Bohr’s fifth orbit for the hydrogen atom is 1.3225 nm.**

**17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. **

**Ans : **The longest wavelength transition in the Balmer series corresponds to the transition from n=2 to n=3. The wavenumber (ṽ) for this transition can be calculated using the Rydberg formula:

**ṽ = R_H * (1/n₁² – 1/n₂²)**

where:

R_H = Rydberg constant (109677 cm⁻¹)

n₁ = initial energy level (2 for Balmer series)

n₂ = final energy level (3 for the longest wavelength transition)

Entering the values into the formula,

ṽ = 109677 cm⁻¹ * (1/2² – 1/3²)

ṽ = 109677 cm⁻¹ * (1/4 – 1/9)

ṽ = 109677 cm⁻¹ * (5/36)

ṽ ≈ 1.523 × 10⁶ m⁻¹

**Hence the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen is approximately 1.523 × 10⁶ m⁻¹.**

**18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.**

**Ans : **To calculate the energy required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and the wavelength of the light emitted when the electron returns to the ground state, we can use the following formulas:

**1. Energy difference between two energy levels:**

ΔE = -R_H * (1/n₂² – 1/n₁²)

where:

ΔE is the energy difference

R_H is the Rydberg constant (2.18 × 10⁻¹⁸ J)

n₁ is the initial energy level

n₂ is the final energy level

**2. Relationship between energy and wavelength:**

E = hc / λ

**Part (i): Energy required to shift the electron**

n₁ = 1 (ground state)

n₂ = 5 (fifth orbit)

ΔE = -2.18 × 10⁻¹⁸ J * (1/5² – 1/1²)

ΔE = -2.18 × 10⁻¹⁸ J * (0.04 – 1)

**ΔE ≈ 2.09 × 10⁻¹⁸ J **

**Part (ii): Wavelength of the emitted light**

When the electron returns to the ground state, it transitions from n = 5 to n = 1.

ΔE

=-2.18 × 10⁻¹⁸ J * (1/1² – 1/5²)

ΔE ≈ 2.04 × 10⁻¹⁸ J

Now, using the equation E = hc / λ, we can solve for λ:

λ = hc / ΔE

λ = (6.626 × 10⁻³⁴ J s) * (3.00 × 10⁸ m/s) / (2.04 × 10⁻¹⁸ J)

**λ ≈ 9.72 × 10⁻⁸ m**

**Therefore, the wavelength of the light emitted when the electron returns to the ground state is approximately 9.72 × 10⁻⁸ m, or 972 Å.**

**19. The electron energy in hydrogen atom is given by En = (–2.18 × 10–18 )/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?**

**Ans : **To calculate the energy required to remove an electron completely from the n = 2 orbit of a hydrogen atom, we need to calculate the ionization energy. The ionization energy is the energy required to move the electron from its current state (n=2) to infinity.

**En = -2.18 × 10⁻¹⁸ / n² J**

Hence, energy of an electron in the n=2 orbit is:

E₂

= -2.18 × 10⁻¹⁸ / 2² J

E₂

= -5.45 × 10⁻¹⁹ J

To ionize the electron from this orbit, we need to provide it with an energy equal to the negative of its current energy. So, the ionization energy is:

Ionization energy = -E₂

Ionization energy = -(-5.45 × 10⁻¹⁹ J)

Ionization energy = 5.45 × 10⁻¹⁹ J

Now, to find the longest wavelength of light that can be used to cause this transition, we can use the relationship between energy and wavelength:

E = hc / λ

λ = hc / E

Substituting the values, we get:

λ = (6.626 × 10⁻³⁴ J s) × (3.00 × 10⁸ m/s) / (5.45 × 10⁻¹⁹ J)

λ ≈ 3.65 × 10⁻⁷ m

Finally, converting the wavelength to centimeters:

λ ≈ 3.65 × 10⁻⁷ m * (100 cm / 1 m)

λ ≈ 3.65 × 10⁻⁵ cm

**Therefore, the longest wavelength of light that can be used to cause this transition is approximately 3.65 × 10⁻⁵ cm.**

**20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1. **

**Ans : **To calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m/s, we can use de Broglie’s equation:

λ = h / mv

λ = (6.626 × 10⁻³⁴ J s) / (9.11 × 10⁻³¹ kg) × (2.05 × 10⁷ m/s)

λ ≈ 3.54 × 10⁻¹¹ m

Therefore, the wavelength of the electron is approximately 3.54 × 10⁻¹¹ meters.

**21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength. **

**Ans : **To calculate the wavelength of an electron

λ = h / mv

However, we need to find the velocity of the electron first, using the kinetic energy (KE):

KE = (1/2)mv²

Rearranging to solve for v:

v = √(2KE / m)

Substituting the values:

v = √(2 * 3.0 × 10⁻²⁵ J / 9.1 × 10⁻³¹ kg)

v ≈ 8.1 × 10⁶ m/s

To find the wavelength:

λ = (6.626 × 10⁻³⁴ J s) / (9.1 × 10⁻³¹ kg) × (8.1 × 10⁶ m/s)

λ ≈ 8.9 × 10⁻¹¹ m

**Therefore, the wavelength of the electron is approximately 8.9 × 10⁻¹¹ meters.**

**22. Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar. **

**Ans : **Isoelectronic species have the same number of electrons.

Let’s analyze the given ions and determine their electron configurations:

Na⁺: 1s² 2s² 2p⁶

(10 electrons)

K⁺: 1s² 2s² 2p⁶ 3s² 3p⁶

(18 electrons)

Mg²⁺: 1s² 2s² 2p⁶

(10 electrons)

Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶

(18 electrons)

S²⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

(18 electrons)

Ar: 1s² 2s² 2p⁶ 3s² 3p⁶

(18 electrons)

From the electron configurations, we can see that:

**Na⁺ and Mg²⁺ both have 10 electrons.**

**K⁺, Ca²⁺, S²⁻, and Ar all have 18 electrons**.

hence, the isoelectronic species are:

**Na⁺ and Mg²⁺**

**K⁺, Ca²⁺, S²⁻, and Ar**

**23.(i)Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2– (d) F–**

** (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? **

**(iii) Which atoms are indicated by the following configurations ? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.**

**Ans : **

**Electronic configurations**

(a) H- (hydride ion)

Neutral hydrogen atom: 1s¹

Gaining an electron: 1s²

(b) Na+ (sodium ion)

Neutral sodium atom: 1s² 2s² 2p⁶ 3s¹

Losing an electron: 1s² 2s² 2p⁶

(c) O²⁻ (oxide ion)

Neutral oxygen atom: 1s² 2s² 2p⁴

Gaining two electrons: 1s² 2s² 2p⁶

(d) F- (fluoride ion)

Neutral fluorine atom: 1s² 2s² 2p⁵

Gaining an electron: 1s² 2s² 2p⁶

**Atomic numbers**

(a) 3s¹ represents the element sodium (Na), which has an atomic number of 11.

(b) 2p³ represents the element nitrogen (N), which has an atomic number of 7.

(c) 3p⁵ represents the element chlorine (Cl), which has an atomic number of 17.

**Atoms which indicate by the following configurations**

(a) [He] 2s¹ represents the element lithium (Li).

(b) [Ne] 3s2 3p3 represents the element phosphorus (P).

(c) [Ar] 4s2 3d1 represents the element scandium (Sc).

**24. What is the lowest value of n that allows g orbitals to exist? **

**Ans : Orbitals are designated by the principal quantum number (n). The possible values for the angular momentum quantum number (l) are 0 to n-1.**

For n = 1, l can be 0 (s orbital)

For n = 2, l can be 0 (s orbital) or 1 (p orbital)

For n = 3,

l can be 0 (s orbital),

1 (p orbital),

or 2 (d orbital)

For n = 4, l can be 0 (s orbital), 1 (p orbital), 2 (d orbital), or 3 (f orbital)

For n = 5, l can be 0 (s orbital), 1 (p orbital), 2 (d orbital), 3 (f orbital), or 4 (g orbital)

**Hence the lowest value is 5.**

**25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.**

**Ans : **The quantum numbers n, l, and ml describe the state of an electron in an atom.

n is the principal quantum number. It determines the energy level of the electron and can have any positive integer value (1, 2, 3, …).

l is the angular momentum quantum number. It determines the shape of the orbital and can have values from 0 to n-1.

ml is the magnetic quantum number. It determines the orientation of the orbital in space and can have values from -l to +l.

For an electron in a 3d orbital:

**n = 3 (because it’s in the 3rd energy level)**

**l = 2 (because d orbitals correspond to l = 2)**

Since l = 2, the possible values of ml are:

ml = -2, -1, 0, 1, 2

Therefore, the possible values of n, l, and ml for an electron in a 3d orbital are:

**n = 3**

**l = 2**

**ml = -2, -1, 0, 1, or 2**

** 26. An atom of an element contains 29 electrons and 35 neutrons. Deduce **

**(i) the number of protons and (ii) the electronic configuration of the element.**

**Ans : (i) Number of protons:**

The atomic number is determined by the number of electrons in a neutral atom & the number of protons in an atom is equal to its atomic number.

In this case, the atom has 29 electrons. Therefore, the atomic number is 29. The number of protons is also 29.

**(ii) Electronic configuration:**

To write the electronic configuration, we need to fill the orbitals in order of increasing energy, following the Aufbau principle, Pauli exclusion principle, and Hund’s rule.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰

**This configuration represents the element copper (Cu), which has an atomic number of 29.**

** 27.Give the number of electrons in the species H****2+****, H****2**** and 0****2+**

**Ans : **To determine the number of electrons in each species, we need to consider the number of electrons in the neutral atoms and any changes due to ionization.

H₂: Hydrogen atoms have 1 electron each. In a molecule of H₂, there are 2 hydrogen atoms, so there are a total of 2 electrons.

H₂⁺: The superscript “+” indicates that the molecule has lost one electron. Therefore, H₂⁺ has 2 – 1 = 1 electron.

O₂: Oxygen atoms have 8 electrons each. In a molecule of O₂, there are 2 oxygen atoms, so there are a total of 16 electrons.

O₂⁺: The superscript “+” indicates that the molecule has lost one electron. Therefore, O₂⁺ has 16 – 1 = 15 electrons.

In summary:

**H₂: 2 electrons**

**H₂⁺: 1 electron**

**O₂: 16 electrons**

**O₂⁺: 15 electrons**

**28 .(i) An atomic orbital has n = 3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f **

**Ans : (i) possible values of l are 0, 1, and 2.**

When l = 0,

the orbital is a 3s orbital.

When l = 1,

the orbital is a 3p orbital.

When l = 2,

the orbital is a 3d orbital.

The possible values of ml range from -l to +l. So for n = 3:

If l = 0,

ml = 0

If l = 1,

ml = -1, 0, or 1

If l = 2,

ml = -2,

-1,

0,

1,

or 2

**(ii) Quantum numbers (ml and l) of electrons for 3d orbital.**

For a 3d orbital:

l = 2

ml = -2,

-1,

0,

1,

or 2

**(iii)Orbitals are possible? 1p, 2s, 2p and 3f**

Possible values of l for a given n are from 0 to n-1.

For n = 1,

l can only be 0

(s orbital)

For n = 2,

l can be 0

(s orbital) or 1 (p orbital)

For n = 3,

l can be 0 (s orbital), 1 (p orbital), or 2 (d orbital)

**Based on this, the following orbitals are possible:**

2s

2p

3d

**The orbital 1p is not possible because for n = 1, l can only be 0. The orbital 3f is not possible because for n = 3, the maximum value of l is 2.**

**29 .****Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3. **

**Ans : **To describe the orbitals using s, p, d notations, we need to relate the principal quantum number (n) and the azimuthal quantum number (l) to the corresponding orbital types.

n = 1, l = 0:

(This represents the 1s orbital.)

n = 3, l = 1:

(This represents the 3p orbital.)

n = 4, l = 2:

( This represents the 4d orbital.)

n = 4, l = 3:

(This represents the 4f orbital.)

Therefore, the orbitals described by the given quantum numbers are:

**(a) 1s**

**(b) 3p**

**(c) 4d**

**(d) 4f**

**30 .Explain, giving reasons, which of the following sets of quantum numbers are not possible. **

** (a) n = 0, l = 0, ml = 0, ms = + ½**

** (b) n = 1, l = 0, ml = 0, ms = – ½ **

**(c) n = 1, l = 1, ml = 0, ms = + ½ **

**(d) n = 2, l = 1, ml = 0, ms = – ½**

**(e) n = 3, l = 3, ml = –3, ms = + ½ **

**f) n = 3, l = 1, ml = 0, ms = + ½ **

**Ans : **The following sets are not possible:

(a) n = 0,

l = 0,

ml = 0,

ms = + ½

(c) n = 1,

l = 1,

ml = 0,

ms = + ½

(e) n = 3,

l = 3,

ml = –3,

ms = + ½

**Reasons:**

(a) n cannot be 0.

Principal quantum number (n) must have be a positive integer.

(c) For n = 1,

the only possible value of l is 0. Having l = 1 is not allowed for n = 1.

(e) For l = 3,

the maximum value of ml is 3. Having ml = -3 is not allowed when l = 3.

**The other sets of quantum numbers (b, d, and f) are all possible combinations.**

**31. How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = – ½ (b) n = 3, l = 0 **

**Ans : (a) n = 4, ms = – ½**

For n = 4,

If l = 0, there is 1 orbital (4s)

If l = 1, there are 3 orbitals (4p)

If l = 2, there are 5 orbitals (4d)

If l = 3, there are 7 orbitals (4f)

Each orbital can hold a maximum of 2 electrons (one with ms = +1/2 and one with ms = -1/2).

**Therefore, the total number of electrons with n = 4 and ms = -1/2 is:**

**1 (from 4s) + 3 (from 4p) + 5 (from 4d) + 7 (from 4f) = 16 electrons**

**(b) n = 3, l = 0**

For n = 3 and l = 0, there is only 1 orbital (3s).

Orbital can hold a maximum 2 electrons.

**Therefore, the maximum number of electrons with n = 3 and l = 0 is 2.**

**32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.**

**Ans : **Showing the Integral Multiple Relationship

Bohr’s Postulate of Angular Momentum:

According to Bohr’s model, the angular momentum of an electron in the nth orbit of a hydrogen atom is quantized and given by:

mvr = nh/2π

where:

m = mass of the electron

v = velocity of the electron

r = radius of the nth Bohr orbit

n = principal quantum number (an integer)

h i= Planck’s constant

**λ = h/p**

For an electron, p = mv.

Relating the Two Equations:

**λ = h/(mv)**

Now, from Bohr’s postulate, we can write:

**mv = nh/2π**

Substituting this into the de Broglie equation:

**λ = h / (nh/2π)**

λ = 2πr/n

Rearranging the equation:

2πr = nλ

**Here, 2πr is the circumference of the nth Bohr orbit.**

**Conclusion:**

**The equation 2πr = nλ shows that the circumference of the nth Bohr orbit is an integral multiple (n) of the de Broglie wavelength associated with the electron revolving around the orbit. This is a fundamental postulate of Bohr’s model, which is consistent with the wave-particle duality of matter.**

** 33 .****What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum ?**

**Ans : **The Rydberg formula for the energy levels of a hydrogen-like atom (such as He+) is:

E_n = -RZ^2 / n^2

where:

energy of the electron in the nth energy level =

E_n

R is the Rydberg constant (2.18 x 10^-18 J)

Z is the atomic number of the atom (Z = 1 for hydrogen, Z = 2 for He+)

n is the principal quantum number

For the Balmer series in He+, the final state is n = 2. The initial state for the transition with the same wavelength in hydrogen would be n = 1.

**Therefore, the transition in the hydrogen spectrum that would have the same wavelength as the Balmer transition n = 4 to n = 2 in He+ is n = 2 to n = 1. **

** 34 . Calculate the energy required for the process He+ (g) He2+ (g) + e– The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom–1.**

**Ans : **We use following formula for calculating ionization energy

**IE = -13.6 * Z² / n² eV**

where:

IE =ionization energy in electron volts (eV)

Z = atomic number of the ion

n = number of the electron

For He⁺, Z = 2 (since it has 2 protons) and n = 1 (ground state). Substituting these values into the formula, we get:

IE(He⁺) = -13.6 * 2² / 1² eV

IE(He⁺) = -54.4 eV

Since the ionization energy is the energy required to remove the electron to infinity, the negative sign indicates that the electron is bound to the nucleus. Therefore, the energy required to remove the electron from He⁺ is 54.4 eV.

To convert this to joules, we can use the conversion factor 1 eV = 1.602 × 10⁻¹⁹ J:

IE(He⁺) = 54.4 eV * 1.602 × 10⁻¹⁹ J/eV

IE(He⁺) ≈ 8.72 × 10⁻¹⁸ J

**Therefore, the energy required for the process He⁺(g) → He²⁺(g) + e⁻ is approximately 8.72 × 10⁻¹⁸ J. **

**35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. **

**Ans : **To calculate the number of carbon atoms that can fit in a straight line across a length of 20 cm, we need to first convert the length of the scale to nanometers and then divide it by the diameter of a carbon atom.

**Converting centimeters to nanometers:**

1 cm = 10⁷ nm

20 cm = 20 * 10⁷ nm = 2 × 10⁸ nm

**Calculating the number of carbon atoms:**

Number of carbon atoms = Length of scale / Diameter of carbon atom

Number of carbon atoms = (2 × 10⁸ nm) / (0.15 nm)

Number of carbon atoms ≈ 1.33 × 10⁹

**Therefore, approximately 1.33 × 10⁹ carbon atoms can be placed side by side in a straight line across a length of 20 cm.**

**36. 2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.**

**Ans : **To calculate the radius of a carbon atom, we need to first find the length occupied by a single carbon atom. Then, we can divide the diameter of the arrangement by the number of carbon atoms to find the diameter of a single carbon atom, and finally, divide the diameter by 2 to get the radius.

**Finding the length occupied by a single carbon atom:**

**Length occupied by one carbon atom = Total length / Number of carbon atoms**

Length occupied by one carbon atom = 2.4 cm / (2 × 10⁸ atoms)

Length occupied by one carbon atom = 1.2 × 10⁻⁸ cm

**Since the diameter of the carbon atom is twice its radius, we can directly say:**

Radius of carbon atom = (1.2 × 10⁻⁸ cm) / 2

Radius of carbon atom = 6.0 × 10⁻⁹ cm

**Therefore, the radius of a carbon atom is 6.0 × 10⁻⁹ cm.**

**37. The diameter of zinc atom is 2.6 Å. Calculate **

**(a) radius of zinc atom in pm and**

** (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.**

**Ans : (a) To convert angstroms (Å) to picometers (pm), we multiply by 100.**

Radius of zinc atom = (2.6 Å) * (100 pm/Å) = 260 pm

**(b) To calculate the number of zinc atoms in a length of 1.6 cm, we need to convert centimeters to angstroms and then divide the length by the diameter of a zinc atom**.

1 cm = 10¹⁰ Å

Length of 1.6 cm in angstroms = 1.6 * 10¹⁰ Å

Number of zinc atoms = (1.6 * 10¹⁰ Å) / (2.6 Å) ≈ 6.15 × 10⁹

**Therefore, the radius of a zinc atom is 260 pm, and approximately 6.15 × 10⁹ zinc atoms can be arranged side by side in a length of 1.6 cm.**

** 38. A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it. **

**Ans : **To calculate the number of electrons present in a particle with a given charge, we can use the following formula:

**Number of electrons =**

** Total charge **

**———————————**

** Charge of a single electron**

Charge of a single electron = -1.602 × 10⁻¹⁹ C.

Substituting the given values into the formula:

Number of electrons = (2.5 × 10⁻¹⁶ C) / (-1.602 × 10⁻¹⁹ C)

Number of electrons ≈ -1560

**Since the number of electrons cannot be negative, we can conclude that the particle has 1560 excess electrons.**

**39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it. **

**Ans : **To calculate the number of electrons present on the oil drop, we can use the following formula:

**Number of electrons = Total charge **

** ———————————**

** Charge of a single electron **

Here, -1.602 × 10⁻¹⁹ C is the charge of a single electron

Substituting the given values into the formula:

Number of electrons = (-1.282 × 10⁻¹⁸ C)

——————————

(-1.602 × 10⁻¹⁹ C)

Number of electrons ≈ 8

**Therefore, there are approximately 8 electrons present on the oil drop.**

**40 .In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ? **

**Ans : **If a thin foil of light atoms like aluminum were used instead of heavy atoms like gold or platinum in Rutherford’s experiment, the following differences would be observed:

**Fewer alpha particles would be deflected:** Lighter atoms have smaller nuclei, which means there is less positive charge concentrated in a smaller area. This would result in fewer alpha particles being deflected at large angles.

**The alpha particles that are deflected would likely experience smaller deflections: **The smaller nucleus of a lighter atom would exert a weaker electrostatic force on the alpha particles, leading to smaller deflections.

**There might be a higher probability of alpha particles passing straight through the foil:** With fewer and smaller nuclei, there would be more space for alpha particles to pass through without being deflected.

**In summary, using a thin foil of light atoms would lead to fewer and smaller deflections of alpha particles, and a higher probability of alpha particles passing straight through. This would provide less evidence for the existence of a small, dense, positively charged nucleus, which was a key finding of Rutherford’s experiment.**

**41. Symbols 35 79Br and 79Br can be written, whereas symbols 79 35Br and 35Br are not acceptable. Answer briefly. **

**Ans : **The correct notation for an isotope are:

A

Z

X

where:

A = mass number

(total number of protons and neutrons)

Z = atomic number

(number of protons)

X = chemical symbol**

Therefore, the symbols 35Br and 79Br are acceptable because they follow this notation. The mass number (A) is written before the chemical symbol (X), and the atomic number (Z) is written as a subscript.

On the other hand, the symbols 79 35Br and 35Br are not acceptable because they do not follow the correct order of the mass number and atomic number. The mass number should always be written before the chemical symbol.

**42 .An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. **

**Ans : **To determine the atomic symbol, we need to find the atomic number (Z) and the mass number (A).

Given:

Mass number (A) = 81

Number of neutrons (N) = 1.317 * Number of protons (Z)

Using the relationship between A, Z, and N:

A = Z + N

Substitute the given expression for N:

81 = Z + 1.317Z

81 = 2.317Z

Z ≈ 35

Therefore, the atomic number (Z) is 35.

Identifying the element:

The element with atomic number =

35 is bromine (Br).

**So, the atomic symbol is 81Br.**

**43. An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion. **

**Ans : Consider the number of electrons as “x”.**

According to the problem, the number of neutrons is 11.1% more than the number of electrons. So, the number of neutrons can be expressed as:

Neutrons = x + 0.111x = 1.111x

Mass number = sum of protons and neutrons

Since the ion has a negative charge, it means it has one more electron than protons. Therefore, the number of protons is x – 1.

So, we can write the equation for mass number as:

Mass number = Protons + Neutrons

37 = (x – 1) + 1.111x

Solving for x:

37 = 2.111x – 1

38 = 2.111x

x ≈ 18

So, the number of electrons is 18. Since the ion has a negative charge, it gained one electron. Therefore, the number of protons is 18 – 1 = 17.

The element with 17 protons is chlorine. Since the ion gained one electron, it has a negative charge.

**Therefore, the symbol of the ion is Cl-**

**44 .An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion. **

**Ans : **Consider the number of electrons as “x”.

According to the problem, the number of neutrons is 30.4% more than the number of electrons. So, the number of neutrons can be expressed as:

Neutrons = x + 0.304x = 1.304x

Mass number = sum of protons and neutrons .

Since the ion has a positive charge of 3 units, it means it has 3 fewer electrons than protons. Therefore, the number of protons is x + 3.

So, we can write the equation for mass number as:

Mass number = Protons + Neutrons

56 = (x + 3) + 1.304x

Solving for x:

56 = 2.304x + 3

53 = 2.304x

x ≈ 23

So, the number of electrons is 23. Since the ion has a positive charge of 3, it lost 3 electrons. Therefore, the number of protons is 23 + 3 = 26.

The element with 26 protons is iron. Since the ion lost 3 electrons, it has a positive charge of 3.

**Therefore, the symbol of the ion is Fe³⁺. **

**45 ****.Arrange the following type of radiations in increasing order of frequency: **

**(a) radiation from microwave oven**

** (b) amber light from traffic signal **

**(c) radiation from FM radio **

**(d) cosmic rays from outer space and **

**(e) X-rays.**

**Ans : FM < microwave < amber color < X-rays < cosmic rays.**

**46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser. **

**Ans : **To calculate the power of the nitrogen laser, we need to find the energy of each photon and then multiply it by the total number of photons.

**1. Find the energy of a single photon:**

We can use the equation:

**E = hc / λ**

where:

E =

energy of a photon

h =

Planck’s constant (6.626 x 10^-34 J s)

c =

speed of light (3 x 10^8 m/s)

λ =

wavelength of the radiation (337.1 nm = 337.1 x 10^-9 m)

Substituting the values:

E = (6.626 x 10^-34 J s) * (3 x 10^8 m/s) / (337.1 x 10^-9 m)

E ≈ 5.87 x 10^-19 J

**2. Calculate the total energy:**

The total energy emitted by the laser is the product of the energy of a single photon and the total number of photons:

**Total energy =**

** Energy of a single photon* Number of photons**

Total energy = (5.87 x 10^-19 J) * (5.6 x 10^24)

Total energy ≈ 3.28 x 10^6 J

**3. Calculate the power:**

Power is the rate of energy transfer. Since the problem doesn’t specify the time taken to emit these photons, we can assume it’s emitted instantaneously. Therefore, power is equal to the total energy.

Power = Total energy

Power ≈ 3.28 x 10^6 J

**Therefore, the power of the nitrogen laser is approximately 3.28 x 10^6 watts.**

**47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, **

** (b) distance traveled by this radiation in 30 s **

** (c) energy of quantum and**

** (d) number of quanta present if it produces 2 J of energy.**

**Ans : (a) Calculate**

** frequency = speed of light / wavelength**

Substituting the values:

frequency = (3 x 10^8 m/s) / (616 x 10^-9 m)

frequency ≈ 4.87 x 10^14 Hz

**(b) To find the distance traveled, we can use the formula:**

**Distance = speed x time**

Substituting the values:

distance = (3 x 10^8 m/s) x (30 s)

distance = 9 x 10^9 m

**(c) To find the energy of a quantum, we can use the formula:**

**Energy = Planck’s constant x frequency**

Substituting the values:

energy = (6.626 x 10^-34 J s) x (4.87 x 10^14 Hz)

energy ≈ 3.23 x 10^-19 J

**(d) To find the number of quanta, we can use the formula:**

**Number of quanta =**

** Total energy **

** ———————————-**

** Energy of a quantum**

Substituting the values:

number of quanta = 2 J / (3.23 x 10^-19 J)

number of quanta ≈ 6.2 x 10^18

Therefore, the answers are:

**(a) The frequency of emission is approximately 4.87 x 10^14 Hz.**

**(b) The radiation travels a distance of approximately 9 x 10^9 meters in 30 seconds.**

**(c) The energy of a quantum is approximately 3.23 x 10^-19 Joules.**

**(d) The number of quanta present if it produces 2 Joules of energy is approximately 6.2 x 10^18.**

**48. ****In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector. **

**Ans : **The time duration is 2 nanoseconds, which is equal to 2 x 10⁻⁹ seconds.

The frequency can be calculated by taking the reciprocal of the time duration: 1 / (2 x 10⁻⁹ seconds) = 5 x 10⁸ s⁻¹.

The energy of one photon is calculated using the formula E = hv, where h is Planck’s constant (6.626 x 10⁻³⁴ Js) and v is the frequency. Plugging in the values, we get: E = (6.626 x 10⁻³⁴ Js) x (5 x 10⁸ s⁻¹) = 3.3125 x 10⁻²⁵ J.

**The number of photons can be calculated by dividing the total energy of the source by the energy of one photon=**

**3.3125 x 10⁻²⁵ J **

**——————————**

**(3.3125 x 10⁻²⁵ J/photon) **

= 2.5 x 10¹⁵ photons.

Finally, the total energy of the source is calculated by multiplying the energy of one photon by the number of photons=

3.3125 x 10⁻²⁵ J/photon x 2.5 x 10¹⁵ photons

**= 8.28 x 10⁻¹⁰ J.**

**49. ****Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.**

**Ans : **The time duration is 2 nanoseconds, which is equal to 2 x 10^-9 seconds.

**The frequency is calculated by taking the reciprocal of the time duration =**

** 1 **

**—————————–**

** (2 x 10^-9 seconds) **

= 5 x 10^8 s^-1

The energy of one photon is calculated using the formula E = hv, where h is Planck’s constant (6.626 x 10^-34 Js) and v is the frequency.

Plugging in the values, we get:

E = (6.626 x 10^-34 Js) x (5 x 10^8 s^-1) = 3.3125 x 10^-25 J.

The number of photons is calculated by dividing the total energy of the source by the energy of one photon=

3.3125 x 10^-25 J

————————————————–

(3.3125 x 10^-25 J/photon)

= 2.5 x 10^15 photons.

Finally, the total energy of the source is calculated by multiplying the energy of one photon by the number of photons=

3.3125 x 10^-25 J/photon x 2.5 x 10^15 photons

**= 8.28 x 10^-10 J.**

**50 .****The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states. **

**Ans : **To calculate the frequency of each transition and the energy difference between the two excited states, we can use the following equations:

**Frequency (ν):**

**ν = c / λ**

where:

c=

speed of light (3 x 10^8 m/s)

λ is the wavelength

**Energy (E):**

**E = hν**

where:

h =

Planck’s constant (6.626 x 10^-34 Js)

Energy difference (ΔE):

ΔE = E2 – E1

where:

E2 is the energy of the excited state with the longer wavelength (589.6 nm)

E1 is the energy of the excited state with the shorter wavelength (589 nm)

**Determine moles of CaCO₃ needed:**- From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of

Calculations:

Frequency for 589 nm:

ν1 = (3 x 10^8 m/s) / (589 x 10^-9 m)

ν1 ≈ 5.093 x 10^14 s^-1

**Frequency for 589.6 nm:**

ν2 = (3 x 10^8 m/s) / (589.6 x 10^-9 m)

ν2 ≈ 5.088 x 10^14 s^-1

**Energy difference:**

ΔE = (6.626 x 10^-34 Js) * (5.093 x 10^14 s^-1 – 5.088 x 10^14 s^-1)

ΔE ≈ 3.31 x 10^-22 J

Therefore:

The frequency of the transition at 589 nm is approximately 5.093 x 10^14 s^-1.

The frequency of the transition at 589.6 nm is approximately 5.088 x 10^14 s^-1.

**The energy difference between the two excited states is approximately 3.31 x 10^-22 Joules.**

**51 .The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.**

**Ans : (a) Threshold wavelength:**

The threshold wavelength is the minimum wavelength of light that can eject an electron from the metal surface. It can be calculated using the formula:

λ₀ = hc / Φ

where:

λ₀ =

the threshold wavelength

h =

Planck’s constant (6.626 x 10^-34 Js)

c =

speed of light (3 x 10^8 m/s)

Φ =

work function (1.9 eV, which is 1.9 x 1.6 x 10^-19 J)

Substituting the values:

λ₀ = (6.626 x 10^-34 Js) * (3 x 10^8 m/s) / (1.9 x 1.6 x 10^-19 J)

**λ₀ ≈ 653 nm**

**(b) Threshold frequency:**

The threshold frequency=

minimum frequency of light can ejects an electron from the metal surface.

It can be calculated using the formula:

ν₀ = c / λ₀

where:

ν₀ =

threshold frequency

c =

speed of light (3 x 10^8 m/s)

λ₀=

threshold wavelength (calculated in part (a))

Substituting the values:

ν₀ = (3 x 10^8 m/s) / (653 x 10^-9 m)

**ν₀ ≈ 4.59 x 10^14 Hz**

**(c) Kinetic energy of the ejected photoelectron:**

Calculate kinetic energy of the ejected photoelectron ,by using following formula

KE = hν – Φ

where:

KE is the kinetic energy

h is Planck’s constant (6.626 x 10^-34 Js)

ν is the frequency of the incident light (500 nm)

Φ is the work function (1.9 eV)

First, we need to convert the wavelength of the incident light to frequency:

ν = c / λ = (3 x 10^8 m/s) / (500 x 10^-9 m) = 6 x 10^14 Hz

Then, we can calculate the kinetic energy:

KE = (6.626 x 10^-34 Js) * (6 x 10^14 Hz) – (1.9 x 1.6 x 10^-19 J)

**KE ≈ 9.31 x 10^-20 J**

**(d) Velocity of the ejected photoelectron:**

Kinetic energy of the photoelectron can be expressed as:

KE = 1/2 mv²

where:

m is the mass of an electron

(9.11 x 10^-31 kg)

v is the velocity of the photoelectron

Solving for v:

v = √(2KE/m)

v = √(2 x 9.31 x 10^-20 J / 9.11 x 10^-31 kg)

v ≈ 4.52 x 10^5 m/s

**Therefore:**

**The threshold wavelength is 653 nm.**

**The threshold frequency is 4.59 x 10^14 Hz. **

**The kinetic energy of the ejected photoelectron is 9.31 x 10^-20 J.**

**The velocity of the ejected photoelectron is 4.52 x 10^5 m/s.**

**52.**** ****Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, **

**(b) Planck’s constant. λ (nm) 500 450 400 v × 10–5 (cm s–1) 2.55 4.35 5.35 **

**Ans : **

The threshold wavelength is denoted by λ₀, which is equal to 450 nm or 4.50 x 10^-9 meters.

According to the photoelectric effect equation, hv – hv₀ = 1/2 mv², where h is Planck’s constant, v is the frequency, v₀ is the threshold frequency, m is the mass of the electron, and velocity stand for v of the ejected electron.

Substituting the results of three experiments into this equation, we get the following three equations:

(1) hc/10^-9 (1/500 – 1/λ₀) = 1/2 m (2.55 x 10^5)²

(2) hc/10^-9 (1/450 – 1/λ₀) = 1/2 m (4.35 x 10^5)²

(3) hc/10^-9 (1/400 – 1/λ₀) = 1/2 m (5.35 x 10^5)²

Dividing equation 3 by equation 1, we get:

(λ₀ – 450) / (λ₀ – 500)

= (4.35 x 10^5)² / (2.99 x 10^5)² x 450 / 500

= 2.619

Solving for λ₀, we get:

**λ₀ = 531 nm**

**53 .The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal. **

**Ans : **The threshold wavelength is 256.7 nm, which is equal to 256.7 x 10^-9 meters.

The energy of a photon is calculated using the equation E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength.

putting in the values, we get:

E = (6.626 x 10^-34 Js) x (3 x 10^8 m/s) / (256.7 x 10^-9 m)

= 7.74 x 10^-18 J

**To convert the energy from joules to electron volts, we multiply by the conversion factor 1.602 x 10^-19 J/eV. This is because one electron volt is equal to 1.602 x 10^-19 joules.**

E = 7.74 x 10^-18 J / (1.602 x 10^-19 J/eV)

= 4.83 eV

The equation E = E₀ + K.E. relates the total energy of the photon (E) to the work function (E₀) and the kinetic energy of the ejected electron (K.E.).

Substituting the known values, we get:

4.83 eV = E₀ + 0.35 eV

**Solving for E₀, we get:**

E₀ = 4.83 eV – 0.35 eV

** = 4.48 eV**

**54. If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus. **

** Ans :**The wavelength is 150 picometers, which is equal to 150 x 10^-12 meters or 1.5 x 10^-10 meters.

The velocity of the electron is 1.5 x 10^7 meters per second.

**To find kinetic energy (K.E.) ,we use formula**

** K.E. = 1/2 mv², **

Where,

m = mass of the electron

and v = velocity.

Putting in the values,

K.E. = 1/2 x 9.1 x 10^-31 kg x (1.5 x 10^7 m/s)²

= 1.02375 x 10^-17 J

The energy of a photon (E) is calculated using the formula E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength.

Putting in the values, we get:

E = (6.626 x 10^-34 Js) x (3 x 10^8 m/s) / (1.5 x 10^-10 m) = 13.252 x 10^-16 J

The work function (E₀) is calculated using the equation E = E₀ + K.E.,

Where,

E = total energy of the photon

K.E. = kinetic energy of the ejected electron.

**Rearranging the equation, we get:**

E₀ = E – K.E. = 13.252 x 10^-16 J – 1.02375 x 10^-16 J

= 12.228 x 10^-16 J

To convert the energy from joules to electron volts, we multiply by the conversion factor 1.602 x 10^-19 J/eV. This is because one electron volt is equal to 1.602 x 10^-19 joules.

E₀ = 12.228 x 10^-16 J / (1.602 x 10^-19 J/eV)

**= 7.63 x 10^3 e**

**55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. **

**Ans : **The Rydberg formula for the energy levels of a hydrogen atom is given by:

**v = (3.29 x 10^15 Hz) * (1/3² – 1/n²)**

where v is the frequency of the emitted radiation, n is the principal quantum number of the final state, and 3 is the principal quantum number of the initial state in this case (Paschen series).

The wavelength λ can be calculated using the formula λ = c/v, where c is the speed of light

Substituting the given values, we get:

λ = 1285 nm

= 1.285 x 10^-6 m

v = (3 x 10^8 m/s / (1.285 x 10^-6 m) = 2.3346 x 10^14 s^-1

**Substituting the values of v and λ into the Rydberg formula, we get:**

2.3346 x 10^14 s^-1 = (3.29 x 10^15 Hz) * (1/3² – 1/n²)

Simplifying the equation, we get:

0.071 = 1/9 – 1/n²

Solving for n, we get:

n² = 1 / (0.071 + 1/9) = 25

**Therefore, n = 5.**

This means that the electron in the **hydrogen atom transitioned from the n = 5 energy level to the n = 3 energy level, **emitting radiation in the Paschen series, which lies in the infrared region of the spectrum.

**56 .Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum. **

**Ans : **The radius of the orbit of a hydrogen-like species is given by the formula:

**r = 0.529 * n² * A / Z**

where r is the radius of the orbit (in pm), n is the principal quantum number, A is the atomic number, and Z is the atomic number of the hydrogen-like species.

For the first orbit n is 1, & radius is:

r₁ = 1322.5 pm / Z

For the second orbit n is 2, & the radius is:

r₂ = 211.6 pm / Z

The ratio of the radii of the first and second orbits is:

r₁ / r₂ = 1322.5 / 211.6 = 6.25

By taking the square root of both sides, we obtain:

n₁ / n₂ = 2.5

Since n₁ = 2, we can solve for n₂:

n₂ = 2.5 * 2 = 5

Therefore, this transition corresponds to **an electron moving from the 5th orbit to the 2nd orbit.** This is a transition in the Balmer series.

The wavenumber (ṽ) can be calculated using the formula

ṽ = 1.097 x 10^7 m^-1 * (1/n₁² – 1/n₂²)

Substituting the values, we get:

ṽ = 1.097 x 10^7 m^-1 * (1/2² – 1/5²)

= 23.037 x 10^5 m^-1

The wavelength (λ) can be calculated using the formula λ = 1/ṽ:

λ = 1 / (23.037 x 10^5 m^-1) = 4.34 x 10^-7 m

** = 434 nm**

**Therefore, the wavelength of the emitted radiation lies in the visible region of the spectrum.**

**57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron. **

**Ans : **we can use the de Broglie equation:

λ = h / mv

where:

λ i=

de Broglie wavelength

h =

Planck’s constant (6.626 x 10^-34 Js)

m =

mass of the electron

(9.11 x 10^-31 kg)

v =

velocity of the electron (1.6 x 10^6 m/s)

Putting the values:

λ =

(6.626 x 10^-34 Js)

——————————————————————

((9.11 x 10^-31 kg) * (1.6 x 10^6 m/s))

λ ≈ 4.55 x 10^-12 m

**Therefore, the de Broglie wavelength associated with the electron in the electron microscope is approximately 4.55 x 10^-12 meters.**

**58 .Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron. **

**Ans : **To calculate the characteristic velocity of the neutron, we can use the de Broglie equation:

**λ = h / mv**

where:

λ =

de Broglie wavelength (800 pm = 800 x 10^-12 m)

h=

Planck’s constant (6.626 x 10^-34 Js)

m=

mass of a neutron (1.675 x 10^-27 kg)

v =

characteristic velocity

Rearranging the equation to solve for v:

v = h / (mλ)

Putting the values:

v = (6.626 x 10^-34 Js)

—————————————————————

((1.675 x 10^-27 kg) * (800 x 10^-12 m))

v ≈ 4.94 x 10^3 m/s

**Therefore, the characteristic velocity associated with the neutron in the neutron diffraction microscope is approximately 4.94 x 10^3 meters per second.**

**59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it. **

**Ans : **we can use the de Broglie equation:

**λ = h / mv**

where:

λ=

de Broglie wavelength

h =

Planck’s constant (6.626 x 10^-34 Js)

m =

the mass of an electron (9.11 x 10^-31 kg)

v =

velocity of the electron (2.19 x 10^6 m/s)

Substituting the values:

λ = (6.626 x 10^-34 Js) / ((9.11 x 10^-31 kg) * (2.19 x 10^6 m/s))

λ ≈ 3.32 x 10^-10 m

**Therefore, the de Broglie wavelength associated with the electron in Bohr’s first orbit is approximately 3.32 x 10^-10 meters.**

**60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity. **

**Ans : **The velocity (u) is 4.37 x 10^9 meters per second and the mass (m) is 0.1 kilograms.

The de Broglie wavelength (λ) is calculated using the formula λ = h/mu, where h is **Planck’s constant (6.626 x 10^-34 kg m² s^-1)**.

Plugging in the values, we get:

λ = (6.626 x 10^-34 kg m² s^-1) / (0.1 kg x 4.37 x 10^9 m/s)

** = 1.516 x 10^-38 meters.**

**61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.**

**Ans :Heisenberg’s Uncertainty Principle**

Heisenberg’s Uncertainty Principle is a fundamental principle of quantum mechanics that states it is impossible to simultaneously determine the exact position and momentum of a particle with absolute precision. In other words, there is a fundamental limit to the accuracy with which these two properties can be measured.

**Given:**

** Uncertainty in position (Δx) = 0.002 nm = 0.002 x 10^-9 m**

Calculation:

We can use the Heisenberg Uncertainty Principle equation:

Δx * Δp ≥ h / 4π

where:

Δx is the uncertainty in position

Δp is the uncertainty in momentum

h is Planck’s constant (6.626 x 10^-34 Js)

**Rearranging the equation to solve for Δp:**

**Δp ≥ h / (4π * Δx)**

Substituting the values:

Δp ≥ (6.626 x 10^-34 Js) / (4π * 0.002 x 10^-9 m)

Δp ≥ 1.32 x 10^-24 kg m/s

Therefore, the uncertainty in the momentum of the electron is at least 1.32 x 10^-24 kg m/s.

Regarding the given momentum of h/4πm × 0.05 nm:

This value represents the momentum of the electron if its position were known with absolute certainty (Δx = 0). However, according to Heisenberg’s Uncertainty Principle, this is impossible. **The uncertainty in its momentum must be at least 1.32 x 10^-24 kg m/s ,when If the position of the electron is measured with an uncertainty of 0.002 nm.**Therefore, defining the momentum of the electron as h/4πm × 0.05 nm is not accurate in this case

**62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: **

**1. n = 4, l = 2, ml = –2 , ms = –1/2 **

**2. n = 3, l = 2, ml = 1 , ms = +1/2 2024-25 structure of atom 73 **

**3. n = 4, l = 1, ml = 0 , ms = +1/2 **

**4. n = 3, l = 2, ml = –2 , ms = –1/2 **

**5. n = 3, l = 1, ml = –1 , ms = +1/2 **

**6. n = 4, l = 1, ml = 0 , ms = +1/2 2.**

**Ans : **To arrange the electrons in order of increasing energy, we need to consider the principal quantum number (n) and the azimuthal quantum number (l). The electron with a lower n value will have lower energy. If the n values are the same, the electron with a lower l value will have lower energy.

**Here are the electrons arranged in order of increasing energy:**

n = 3, l = 1, ml = -1, ms = +1/2 (3p orbital)

n = 3,

l = 2,

ml = -2,

ms = -1/2 (3d orbital)

n = 3, l = 2, ml = 1, ms = +1/2 (3d orbital)

n = 4, l = 1, ml = 0, ms = +1/2 (4p orbital)

n = 4, l = 1, ml = 0, ms = +1/2 (4p orbital)

n = 4, l = 2, ml = -2, ms = -1/2 (4d orbital)

Note:

Electrons with the same n and l values have the same energy level. This is known as degeneracy.

In this case, electrons 3 and 6 have the same energy because they both occupy 4p orbitals.

Electrons with higher n values have higher energy levels.

Therefore, the order of increasing energy is:

**(3p) < (3d) = (3d) < (4p) = (4p) < (4d)**

**hence, 5 < 2 = 4 < 3 = 6 < 1**

**63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ? **

**Ans : **The effective nuclear charge experienced by an electron is the net positive charge that it feels from the nucleus, after taking into account the shielding effect of other electrons.

In the case of bromine, the 2p, 3p, and 4p electrons are all shielded by the 1s, 2s, and 3s electrons. However, the 4p electrons experience the lowest effective nuclear charge because they are the furthest away from the nucleus and are shielded by the most inner electrons.

**Therefore, the 4p electrons in bromine experience the lowest effective nuclear charge.**

**64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.**

**Ans : **The effective nuclear charge experienced by an electron is the net positive charge that it feels from the nucleus, after taking into account the shielding effect of other electrons.

Here’s a comparison of the effective nuclear charge experienced by electrons in the given pairs of orbitals:

**(i) 2s and 3s:**

The 2s orbital is closer to the nucleus rather than the 3s orbital.

Therefore, the 2s electron experiences a larger effective nuclear charge due to less shielding from inner electrons.

**(ii) 4d and 4f:**

The 4d orbital is closer to the nucleus than the 4f orbital.

Therefore, the 4d electron experiences a larger effective nuclear charge due to less shielding from inner electrons.

**(iii) 3d and 3p:**

The 3d orbital has a more complex shape than the 3p orbital, and its electrons penetrate closer to the nucleus.

Therefore, the 3d electron experiences a slightly larger effective nuclear charge than the 3p electron.

In summary, the orbital that experiences the larger effective nuclear charge is the one that is:

Closer to the nucleus

Has a more penetrating shape

Therefore, among the given pairs, the orbitals experiencing larger effective nuclear charge are:

**2s compared to 3s**

**4d compared to 4f**

**3d compared to 3p**

**65 .The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ? **

**Ans : **The two elements have the following electron configurations:

Aluminum (Al, Z = 13): [Ne] 3s² 3p¹

Silicon (Si, Z = 14): [Ne] 3s² 3p²

Silicon, with a higher atomic number, exerts a more effective nuclear charge on its unpaired electrons than aluminum This means that the nucleus of silicon has a greater positive charge, which attracts the electrons more strongly.

.

**66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr. **

**Ans : Given elements presents electron configurations **

**P =**

1s² ,2s², 2p⁶ ,3s², 3p³

**Si= **

1s² ,

2s² ,

2p⁶ ,

3s² ,

3p²

**Cr=**

1s²,

2s² ,

2p⁶,

3s²,

3p⁶ ,

4s¹,

3d⁵

**Fe=**

1s²

,2s²

,2p⁶

,3s² ,

3p⁶ ,

4s²,

3d⁶

**Kr=**

1s²,

2s² ,

2p⁶,

3s² ,

3p⁶ ,

4s²,

3d¹⁰,

4p⁶

**Now, let’s analyze the valence shells (outermost shells) to determine the number of unpaired electrons:**

P: The 3p subshell has 3 electrons, and each orbital can hold a maximum of 2 electrons. Therefore, there are 3 unpaired electrons in phosphorus.

Si: The 3p subshell has 2 electrons, which can occupy two different orbitals. Therefore, there are 2 unpaired electrons in silicon.

Cr: The 3d subshell has 5 electrons. However, due to the half-filled subshell stability, one electron from the 4s orbital moves to the 3d subshell to create a half-filled configuration. This results in 6 unpaired electrons in chromium.

Fe: The 3d subshell has 6 electrons. In iron, there are 4 unpaired electrons due to the pairing of two electrons in one of the 3d orbitals.

Kr: Krypton has a completely filled valence shell (4p⁶). Therefore, there are 0 unpaired electrons in krypton.

In summary, the number of unpaired electrons in the given elements are:

**P: 3**

**Si: 2**

**Cr: 6**

**Fe: 4**

**Kr: 0**

**67 .(a) How many subshells are associated with n = 4 ? **

** (b) How many electrons will be present in the subshells having ms value of –1/2 for n = 4 ?**

**Ans : **

**(a)**The number of subshells associated with a given principal quantum number (n) is equal to n.

Hence, n is 4, there are 4 subshells.

These subshells are:

4s

4p

4d

4f

** (b)The maximum number of electrons in a subshell is given by the formula 2(2l + 1), where l is the azimuthal quantum number for that subshell.**

For n = 4, the possible values of l are 0, 1, 2, and 3, corresponding to the s, p, d, and f subshells, respectively.

For the 4s subshell (l = 0), there are 2(2*0 + 1) = 2 electrons.

For the 4p subshell (l = 1), there are 2(2*1 + 1) = 6 electrons.

For the 4d subshell (l = 2), there are 2(2*2 + 1) = 10 electrons.

For the 4f subshell (l = 3), there are 2(2*3 + 1) = 14 electrons.

Since each electron can have either +1/2 or -1/2 as its spin quantum number (ms), half of the electrons in each subshell will have ms = -1/2.

**Therefore, the total number of electrons with ms = -1/2 for n = 4 is:**

**2 + 3 + 5 + 7 = 17 electrons.**