Question

The energy gap of an LED is 2.4 eV. When the LED is switched ‘ON’, the momentum of the emitted photons is

• A. 2.56 × 10^-27 kg.m.s^-1
• B. 1.28 × 10^-11 kg.m.s^-1
• C. 0.64 × 10^-27 kg.m.s^-1
• D. 1.28 × 10^-27 kg.m.s^-1

Answer 1

The Correct Option is D

The energy gap of an LED is given as 2.4 eV. The momentum of the emitted photons can be calculated using the following relation:

Momentum (p) = √(2 * m * E)

Where:  
m is the mass of the photon, and 
E is the energy of the photon.

Using the relation between energy and momentum for a photon, we know that:

E = p * c

Where E is the energy of the photon, p is the momentum, and c is the speed of light (approximately 3 × 10⁸ m/s). Therefore, we can rearrange the equation to solve for momentum:

p = E / c

Substituting the values: 
Energy of the photon = 2.4 eV = 2.4 × 1.602 × 10⁻¹⁹ J 
Speed of light c = 3 × 10⁸ m/s

Thus, the momentum of the emitted photons is:

p = (2.4 × 1.602 × 10⁻¹⁹) / (3 × 10⁸)

This gives the momentum as 1.28 × 10⁻²⁷ kg·m/s.

Therefore, the correct answer is D: 1.28 × 10⁻²⁷ kg·m/s.

Answer 2

The energy gap \( E_g \) of the LED is given as 2.4 eV. The energy of a photon is related to its momentum using the equation: \[ E = p \cdot c \] where \( E \) is the energy of the photon, \( p \) is the momentum, and \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \). First, convert the energy from electron volts to joules: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \] \[ E = 2.4 \, \text{eV} = 2.4 \times 1.6 \times 10^{-19} \, \text{J} = 3.84 \times 10^{-19} \, \text{J} \] Now, calculate the momentum using the equation \( E = p \cdot c \): \[ p = \frac{E}{c} = \frac{3.84 \times 10^{-19}}{3 \times 10^8} = 1.28 \times 10^{-27} \, \text{kg} \cdot \text{m/s} \] Thus, the momentum of the emitted photons is \( 1.28 \times 10^{-27} \, \text{kg} \cdot \text{m/s} \).

\(\textbf{Correct Answer:}\) \((D)\) \(1.28 \times 10^{-27} \, \text{kg} \cdot \text{m/s}\)