Class 12 - Unit # 26 : Atomic Physics - Solved Numericals

 

1. Energy of Electron in Bohr Orbit

Calculate the energy of an electron in the n = 2 orbit of a hydrogen atom according to the Bohr model.

Data:

• Number of Orbit (n) = 2
• Mass of Electron (m) = 9.11 × 10⁻³¹ kg
• Charge on Electron (e) = 1.6 × 10⁻¹⁹ C
• Permittivity of Free Space (ε₀) = 8.85 × 10⁻¹² C²/N·m²
• Planck's Constant (h) = 6.63 × 10⁻³⁴ J·s
• Energy of Electron (E₂) = ?

Solution:

According to Bohr's Atomic Model, the energy of an electron in the n-th orbit is given by: E_n = -13.6 / n² eV

For n = 2 orbit: E₂ = -13.6 / 2²

E₂ = -13.6 / 4 = -3.4 eV

In Joules: E₂ = -3.4 × 1.6 × 10⁻¹⁹ J = -5.44 × 10⁻¹⁹ J

Result: The energy of an electron in the n = 2 orbit of a hydrogen atom is -3.4 eV (or -5.44 × 10⁻¹⁹ J).

2. Speed of Electron in Bohr Orbits

Calculate the speed of the electron if it orbits in (a) the smallest allowed orbit and (b) the second smallest orbit? (c) If the electron moves to larger orbits, does its speed increase, decrease, or stay the same?

Data:

• (a) Smallest allowed orbit: n = 1, Speed (v₁) = ?
• (b) Second smallest orbit: n = 2, Speed (v₂) = ?
• (c) Effect of speed in larger orbits = ?

Solution:

According to Bohr's Atomic Model, the orbital velocity is: v_n = e² / (2 × ε₀ × h × n)

Using ground state constants, the speed can be simplified to: v_n = (2.18 × 10⁶) / n m/s

(a) For the smallest allowed orbit (n = 1):

v₁ = (2.18 × 10⁶) / 1 = 2.18 × 10⁶ m/s

(b) For the second smallest orbit (n = 2):

v₂ = (2.18 × 10⁶) / 2 = 1.09 × 10⁶ m/s

(c) Analysis of speed for larger orbits:

Since velocity v is inversely proportional to the orbit number n, the speed decreases as the orbit number or radius increases.

Result: The speed of the electron in (a) the smallest allowed orbit is 2.18 × 10⁶ m/s, (b) the second smallest orbit is 1.09 × 10⁶ m/s, and (c) the speed of the electron decreases with the increase in radius.

3. Photon Emission Transition (n = 3 to n = 1)

What are the (a) energy, (b) magnitude of the momentum, and (c) wavelength of the photon emitted when a hydrogen atom undergoes a transition from a state with n = 3 to a state with n = 1?

Data:

• Initial Orbit (n_i) = 3
• Final Orbit (n_f) = 1
• Rydberg's Constant (R) = 1.097 × 10⁷ m⁻¹
• (a) Energy of Photon (E) = ?
• (b) Momentum (p) = ?
• (c) Wavelength of Photon (λ) = ?

Solution:

(c) Wavelength calculation via Rydberg Formula: 1/λ = R × (1/n_f² - 1/n_i²)

1/λ = 1.097 × 10⁷ × (1/1² - 1/3²)

1/λ = 1.097 × 10⁷ × (1 - 1/9) = 1.097 × 10⁷ × (8/9)

1/λ ≈ 9.751 × 10⁶ m⁻¹

λ = 1 / (9.751 × 10⁶) ≈ 1.026 × 10⁻⁷ m = 102.6 nm

(a) Energy calculation using Planck's Law: E = hc / λ

E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.026 × 10⁻⁷) ≈ 1.938 × 10⁻¹⁸ J

In electron volts: E = (1.938 × 10⁻¹⁸) / (1.6 × 10⁻¹⁹) ≈ 12.11 eV

(b) Momentum calculation using de Broglie's Hypothesis: p = h / λ

p = (6.63 × 10⁻³⁴) / (1.026 × 10⁻⁷)

p ≈ 6.46 × 10⁻² kg·m/s

Result: (a) Energy is equal to 12.11 eV, (b) magnitude of the momentum is 6.46 × 10⁻² kg·m/s, and (c) wavelength of the photon is 102.6 nm.

4. Photon Energy in Transition (n = 5 to n = 2)

What is the energy of the photon emitted by a hydrogen atom when the hydrogen atom changes directly from the n = 5 state to the n = 2 state?

Data:

• Initial Orbit (n_i) = 5
• Final Orbit (n_f) = 2
• Rydberg's Constant (R) = 1.097 × 10⁷ m⁻¹
• Energy of Photon (E) = ?

Solution:

The wavelength of the photon is determined by: 1/λ = R × (1/n_f² - 1/n_i²)

1/λ = 1.097 × 10⁷ × (1/2² - 1/5²)

1/λ = 1.097 × 10⁷ × (1/4 - 1/25) = 1.097 × 10⁷ × (21/100)

1/λ = 2.3037 × 10⁶ m⁻¹ ⇒ λ ≈ 4.34 × 10⁻⁷ m = 434 nm

Using Planck's Law to calculate Energy: E = hc / λ

E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (4.34 × 10⁻⁷) ≈ 4.583 × 10⁻¹⁹ J

Converting into electron volts: E = (4.583 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) ≈ 2.86 eV

Result: The energy of the photon emitted by the hydrogen atom is 2.86 eV.

5. Ionization Work for Hydrogen Atom

How much work must be done to pull apart the electron and the proton that make up the hydrogen atom if the atom is initially in (a) its ground state and (b) the state with n = 3?

Data:

• (a) Initial Orbit (n_i) = 1, Final Orbit (n_f) = ∞
• (b) Initial Orbit (n_i) = 3, Final Orbit (n_f) = ∞
• Work Required / Energy (W) = ?

Solution:

To completely separate the particles, the final orbit state energy targets infinity (E_∞ = 0 eV).

The work required is given by the ionization energy formula: W = E_f - E_i = 0 - (-13.6 / n²) = 13.6 / n² eV

(a) For Ground State (n = 1):

W₁ = 13.6 / 1² = 13.6 eV

(b) For Excited State (n = 3):

W₃ = 13.6 / 3² = 13.6 / 9 ≈ 1.51 eV

Result: The work that must be done is (a) 13.6 eV for the ground state and (b) 1.51 eV for the n = 3 state.

6. Balmer Series Limit and Wavelengths

(a) What is the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines? (b) What is the wavelength of the series limit?

Data:

• Balmer Series Final Orbit (n_f) = 2
• (a) Least energetic photon transition: Initial Orbit (n_i) = 3
• (b) Series limit maximum energy transition: Initial Orbit (n_i) = ∞
• Rydberg's Constant (R) = 1.097 × 10⁷ m⁻¹

Solution:

(a) For the least energetic photon (n = 3 to n = 2):

1/λ = R × (1/2² - 1/3²)

1/λ = 1.097 × 10⁷ × (1/4 - 1/9) = 1.097 × 10⁷ × (5/36)

1/λ = 1.5236 × 10⁶ m⁻¹ ⇒ λ = 1 / (1.5236 × 10⁶) ≈ 6.563 × 10⁻⁷ m = 656.3 nm

(b) For the series limit photon (n = ∞ to n = 2):

1/λ = R × (1/2² - 1/∞²)

1/λ = 1.097 × 10⁷ × (1/4 - 0) = 1.097 × 10⁷ / 4

1/λ = 2.7425 × 10⁶ m⁻¹ ⇒ λ = 1 / (2.7425 × 10⁶) ≈ 3.646 × 10⁻⁷ m = 364.6 nm

Result: (a) The wavelength of light for the least energetic photon emitted in the Balmer series is 656.3 nm and (b) the wavelength of the series limit is 364.6 nm.

7. Laser Photon Energy Calculation

A laser emits light with a wavelength of 632.8 nm and has a power output of 55 mW. Calculate the energy of one photon emitted by this laser.

Data:

• Wavelength (λ) = 632.8 nm = 632.8 × 10⁻⁹ m
• Power Output (P) = 55 mW = 55 × 10⁻³ W
• Energy of Photon (E) = ?

Solution:

According to Planck's Radiation Law, the energy of a single photon is completely determined by wavelength: E = hc / λ

E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (632.8 × 10⁻⁹)

E = (1.989 × 10⁻²⁵) / (632.8 × 10⁻⁹)

E ≈ 3.143 × 10⁻¹⁹ J

Converting into electron volts: E = (3.143 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) ≈ 1.96 eV

Result: The energy of one photon emitted by this laser is 3.143 × 10⁻¹⁹ J (or 1.96 eV).

8. Wavelength of X-ray Photon

Calculate the wavelength of X-rays if the energy of one photon emitted by the X-ray machine is 1.9878 × 10⁻¹⁵ Joules.

Data:

• Energy of Photon (E) = 1.9878 × 10⁻¹⁵ J
• Planck's Constant (h) = 6.63 × 10⁻³⁴ J·s
• Speed of Light (c) = 3 × 10⁸ m/s
• Wavelength (λ) = ?

Solution:

According to Planck's Law: E = hc / λ

Rearranging the formula to isolate the target wavelength variable: λ = hc / E

λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.9878 × 10⁻¹⁵)

λ = (1.989 × 10⁻²⁵) / (1.9878 × 10⁻¹⁵)

λ ≈ 1.0006 × 10⁻¹⁰ m = 0.1 nm = 1.0 Å

Result: The wavelength of the X-rays is 1.0 × 10⁻¹⁰ m (or 0.1 nm).