Class 12 - Unit # 25 : Quantum Physics - Solved Numericals

1. Solar Radiation and Power

The Sun's surface temperature is 5700 K. (i) How much power is radiated by the Sun? (ii) Given that the distance to Earth is about 200 Sun radii, what is the maximum power possible from one square kilometer solar energy installation? (iii) What is the wavelength of maximum intensity of solar radiation?

Data:

• Temperature (T) = 5700 K
• Distance to Earth (r) = 200 R_s
• Installation Area (A) = 1 km² = 10⁶ m²
• (i) Total Radiated Power (P) = ?
• (ii) Maximum Power for Installation (P_max) = ?
• (iii) Wavelength for Max Intensity (λ_max) = ?

Solution:

(i) According to Stefan-Boltzmann Law, the total power radiated by the Sun is: P = σ × T⁴ × 4πR_s²

P = (5.67 × 10⁻⁸) × 5700⁴ × 4πR_s²

P ≈ 7.52 × 10⁸ R_s² W

(ii) The solar radiation intensity at the distance of the Earth is: I_E = P / (4πr²)

I_E = (7.52 × 10⁸ R_s²) / (4π × (200 R_s)²)

I_E = (7.52 × 10⁸) / (4 × 40000) ≈ 1496 W/m²

The power collected by 1 km² (10⁶ m²) installation area is: P_max = I_E × A

P_max = 1496 × 10⁶ = 1.496 × 10⁹ W

(iii) According to Wien's Displacement Law: λ_max = (2.898 × 10⁻³) / T

λ_max = (2.898 × 10⁻³) / 5700

λ_max ≈ 5.08 × 10⁻⁷ m = 508 nm

Result: (i) The power radiated by the Sun is 7.52 × 10⁸ R_s² W, (ii) the maximum power from a 1 km² installation is 1.496 × 10⁹ W, and (iii) the peak wavelength is 508 nm.

2. Peak Thermal Radiation from Skin

The temperature of your skin is approximately 32 °C. What is the wavelength at which the peak occurs in the radiation emitted from your skin?

Data:

• Temperature (T) = 32 °C = 32 + 273.15 = 305.15 K
• Wien's Constant (b) = 2.898 × 10⁻³ m·K
• Wavelength for Max Intensity (λ_max) = ?

Solution:

According to Wien's Displacement Law: λ_max = b / T

λ_max = (2.898 × 10⁻³) / 305.15

λ_max ≈ 9.50 × 10⁻⁶ m = 9.50 µm

Result: The wavelength at which the peak occurs in the radiation emitted from skin is 9.50 µm.

3. Photon Emission Rate

An FM radio transmitter has a power output of 100 kW and operates at a frequency of 94 MHz. How many photons per second does the transmitter emit?

Data:

• Power Output (P) = 100 kW = 10⁵ W
• Frequency (f) = 94 MHz = 94 × 10⁶ Hz
• Time interval (t) = 1.0 s
• Number of Photons (n) = ?

Solution:

According to Planck's Law, the energy of a single photon is: E = h × f

E = (6.63 × 10⁻³⁴) × (94 × 10⁶) = 6.232 × 10⁻²⁶ J

By definition, power is total energy emitted per second: P = (n × E) / t

Rearranging to find the photon count per second (t = 1 s): n = P / E

n = 10⁵ / (6.232 × 10⁻²⁶)

n ≈ 1.61 × 10³⁰

Result: The transmitter emits 1.61 × 10³⁰ photons per second.

4. Work Function via Dual Wavelengths

A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.0 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 4.0 eV. Determine the work function of the metal.

Data:

• Max K.E. 1 (K₁) = 1.0 eV
• Max K.E. 2 (K₂) = 4.0 eV
• Given Condition: λ₂ = λ₁ / 2
• Work Function (ϕ) = ?

Solution:

According to Einstein's Photoelectric Equation: K = (hc / λ) - ϕ

For the 1st Light source: 1.0 = (hc / λ₁) - ϕ ⇒ hc / λ₁ = 1.0 + ϕ

For the 2nd Light source: 4.0 = (hc / λ₂) - ϕ

Substitute λ₂ = λ₁ / 2: 4.0 = 2(hc / λ₁) - ϕ

Substitute the value of (hc / λ₁) from the first step: 4.0 = 2(1.0 + ϕ) - ϕ

4.0 = 2.0 + 2ϕ - ϕ

4.0 = 2.0 + ϕ ⇒ ϕ = 4.0 - 2.0 = 2.0 eV

Result: The work function of the metal is 2.0 eV.

5. Photoelectric Effect on Calcium

A 430 nm violet light is incident on a calcium photoelectrode with a work function of 2.71 eV. Find the energy of the incident photons and the maximum kinetic energy of ejected electrons.

Data:

• Wavelength (λ) = 430 nm = 430 × 10⁻⁹ m
• Work Function (ϕ) = 2.71 eV
• Energy of Photon (E) = ?
• Maximum Kinetic Energy (K_max) = ?

Solution:

According to Planck's Law, the energy of the incident photon is: E = hc / λ

E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (430 × 10⁻⁹)

E = 4.625 × 10⁻¹⁹ J

Converting into electron volts: E = (4.625 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) = 2.89 eV

According to Einstein's Photoelectric Equation: K_max = E - ϕ

K_max = 2.89 - 2.71 = 0.18 eV

Result: The energy of the incident photons is 2.89 eV and the maximum kinetic energy of the ejected electrons is 0.18 eV.

6. Planck's Constant Experimental Estimation

Cut-off frequency for the photoelectric effect in some materials is 8 × 10¹³ Hz. When the incident light has a frequency of 1.2 × 10¹⁴ Hz, the stopping potential is measured as 0.16 V. Estimate a value of Planck's constant from these data and determine the percentage error of your estimation.

Data:

• Threshold Frequency (f₀) = 8 × 10¹³ Hz
• Frequency of Radiation (f) = 1.2 × 10¹⁴ Hz
• Stopping Potential (V_s) = 0.16 V
• Estimated Planck's Constant (h_est) = ?
• Percentage Error = ?

Solution:

According to Einstein's Photoelectric Equation: e × V_s = h(f - f₀)

Rearranging to isolate h: h = (e × V_s) / (f - f₀)

h = (1.6 × 10⁻¹⁹ × 0.16) / (1.2 × 10¹⁴ - 0.8 × 10¹⁴)

h = (2.56 × 10⁻²⁰) / (4.0 × 10¹³)

h = 6.40 × 10⁻³⁴ J·s

Calculating percentage error relative to the standard value (6.63 × 10⁻³⁴ J·s):

Percentage Error = [|6.63 - 6.40| / 6.63] × 100% ≈ 3.47%

Result: The estimated value of Planck's constant is 6.40 × 10⁻³⁴ J·s and the percentage error of estimation is 3.47%.

7. Threshold Evaluation for Listed Metals

The work functions of some metals are given. Determine the number of metals that will show the photoelectric effect when light of 300 nm wavelength falls on them.

Data:

• Wavelength (λ) = 300 nm = 300 × 10⁻⁹ m
• Energy of Photon (E) = ?
• Condition for emission: E > ϕ

Solution:

According to Planck's Law: E = hc / λ

E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (300 × 10⁻⁹)

E = 6.63 × 10⁻¹⁹ J

Converting into electron volts: E = (6.63 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) ≈ 4.14 eV

For the photoelectric effect to take place, the energy of the incident radiation must be greater than the work function of the metal (E > ϕ).

Comparing 4.14 eV to typical alkaline elements: Li (2.9 eV), Na (2.3 eV), K (2.0 eV), and Mg (3.7 eV) are all below 4.14 eV.

Result: 4 metals (Li, Na, K, and Mg) have work functions less than 4.14 eV and will show the photoelectric effect.

8. Compton Scattering Analysis

X-rays with an energy of 300 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 30° relative to the incident X-rays, determine the Compton shift at this angle, the energy of the scattered X-rays, and the energy of the recoiling electron.

Data:

• Incident Photon Energy (E) = 300 keV = 300,000 eV
• Scattering Angle (θ) = 30°
• (i) Compton Shift (Δλ) = ?
• (ii) Scattered X-ray Energy (E') = ?
• (iii) Recoil Electron Kinetic Energy (K_e) = ?

Solution:

(i) The equation for Compton Shift is: Δλ = (h / (m₀c)) × (1 - cos θ)

Δλ = (2.43 × 10⁻¹²) × (1 - cos 30°)

Δλ = (2.43 × 10⁻¹²) × (1 - 0.866)

Δλ = 2.43 × 10⁻¹² × 0.134 = 3.26 × 10⁻¹³ m

(ii) Find the initial wavelength λ: λ = hc / E

λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (300,000 × 1.6 × 10⁻¹⁹) = 4.144 × 10⁻¹² m

The scattered wavelength is: λ' = λ + Δλ

λ' = 4.144 × 10⁻¹² + 0.326 × 10⁻¹² = 4.470 × 10⁻¹² m

The energy of the scattered photon is: E' = hc / λ'

E' = (6.63 × 10⁻³⁴ × 3 × 10⁸) / 4.470 × 10⁻¹² = 4.450 × 10⁻¹⁴ J

In electron volts: E' = (4.450 × 10⁻¹⁴) / (1.6 × 10⁻¹⁹) ≈ 278,125 eV = 278.1 keV

(iii) The kinetic energy of the recoiling electron is given by: K_e = E - E'

K_e = 300 keV - 278.1 keV = 21.9 keV

Result: The Compton shift is 3.26 × 10⁻¹³ m, the scattered X-ray energy is 278.1 keV, and the recoil electron energy is 21.9 keV.

9. Complete Collision Mechanics

A photon with a wavelength of 6.0 × 10⁻¹² m collides with an electron. After the collision the photon wavelength is found to have changed by exactly one Compton wavelength (2.43 × 10⁻¹² m). (i) What is the photon's wavelength after collision? (ii) Through what angle has it been deflected? (iii) What is the trajectory angle for the electron? (iv) What is the electron's kinetic energy in eV?

Data:

• Incident Wavelength (λ) = 6.0 × 10⁻¹² m
• Compton Shift (Δλ) = 2.43 × 10⁻¹² m
• (i) Scattered Wavelength (λ') = ?
• (ii) Scattering Angle (θ) = ?
• (iii) Recoil Electron Angle (ϕ) = ?
• (iv) Electron Kinetic Energy (K_e) = ?

Solution:

(i) The scattered photon wavelength is: λ' = λ + Δλ

λ' = 6.0 × 10⁻¹² + 2.43 × 10⁻¹² = 8.43 × 10⁻¹² m

(ii) According to the definition of Compton Shift: Δλ = λ_c × (1 - cos θ)

Since Δλ = λ_c: 1 = 1 - cos θ ⇒ cos θ = 0 ⇒ θ = 90°

(iii) From conservation equations, the electron recoil angle relationship resolves to: tan ϕ = λ / λ'

tan ϕ = (6.0 × 10⁻¹²) / (8.43 × 10⁻¹²) ≈ 0.7117 ⇒ ϕ ≈ 35.4°

(iv) The kinetic energy of the electron is: K_e = hc(1/λ - 1/λ')

Initial Photon Energy: E = 206.67 keV | Scattered Photon Energy: E' = 147.10 keV

K_e = 206.67 - 147.10 = 59.57 keV = 59570 eV

Result: (i) The wavelength after collision is 8.43 × 10⁻¹² m, (ii) the deflection angle is 90°, (iii) the electron recoil angle is 35.4°, and (iv) the electron K.E. is 59.57 keV.

10. Ground State de Broglie Wavelength

Find the de Broglie wavelength of an electron in the ground state of hydrogen.

Data:

• Ground State Orbit (n) = 1
• Bohr Radius of 1st Orbit (r₁) = 0.529 × 10⁻¹⁰ m
• de Broglie Wavelength (λ) = ?

Solution:

According to Bohr's Atomic Model quantization condition: 2πr = nλ

For the ground state (n = 1), this simplifies to: λ = 2π × r₁

λ = 2 × 3.1416 × (0.529 × 10⁻¹⁰)

λ = 3.324 × 10⁻¹⁰ m = 0.332 nm

Result: The de Broglie wavelength of an electron in the ground state of hydrogen is 0.332 nm.

11. Quantum Uncertainty Limits

Determine the minimum uncertainties in the positions of the following objects if their speeds are known with a precision of 1.0 × 10⁻³ m/s: (a) an electron and (b) a bowling ball of mass 6.0 kg.

Data:

• Precision of speed (Δv) = 1.0 × 10⁻³ m/s
• (a) Mass of electron (m_e) = 9.11 × 10⁻³¹ kg
• (b) Mass of bowling ball (m_b) = 6.0 kg
• Minimum position uncertainty (Δx) = ?

Solution:

According to Heisenberg's Uncertainty Principle: Δx × Δp ≥ h / (4π)

Since Δp = m × Δv, the expression expands to: Δx ≥ h / (4π × m × Δv)

(a) For the Electron:

Δx ≥ (6.63 × 10⁻³⁴) / (4 × 3.1416 × 9.11 × 10⁻³¹ × 1.0 × 10⁻³)

Δx ≥ (6.63 × 10⁻³⁴) / (1.1448 × 10⁻³²) ≈ 5.79 × 10⁻² m = 5.79 cm

(b) For the Bowling Ball:

Δx ≥ (6.63 × 10⁻³⁴) / (4 × 3.1416 × 6.0 × 1.0 × 10⁻³)

Δx ≥ (6.63 × 10⁻³⁴) / (7.5398 × 10⁻²) ≈ 8.79 × 10⁻³³ m

Result: The minimum position uncertainty for (a) the electron is 5.79 cm, and for (b) the bowling ball is 8.79 × 10⁻³³ m.