Question

A transmitter of power 10 kW emits radio waves of wavelength 500 m. The number of photons emitted per second by the transmitter is of the order of:

• A. \( 10^{37} \)
• B. \( 10^{31} \)
• C. \( 10^{25} \)
• D. \( 10^{43} \)

Hint:

The number of emitted photons per second is given by \( N = \frac{P}{E} \), where \( P \) is power and \( E = \frac{hc}{\lambda} \) is the energy of each photon.

Answer 1

The Correct Option is B

Step 1: Use Power and Energy Relation The number of photons emitted per second is given by: \[ N = \frac{P}{E} \] where the energy of a photon is: \[ E = \frac{hc}{\lambda} \] Step 2: Compute Photon Energy \[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{500} \] \[ E = \frac{1.9878 \times 10^{-25}}{500} \] \[ E = 3.9756 \times 10^{-28} \text{ J} \] Step 3: Compute Number of Photons Per Second \[ N = \frac{10^4}{3.9756 \times 10^{-28}} \] \[ N \approx 2.5 \times 10^{31} \] Thus, the correct answer is \( 10^{31} \).

Answer 2

Given:
- Power of transmitter, \( P = 10 \, \text{kW} = 10 \times 10^{3} = 10^{4} \, \text{W} \)
- Wavelength of radio waves, \( \lambda = 500 \, \text{m} \)
- Planck’s constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{m/s} \)

Step 1: Calculate the energy of one photon emitted:
\[ E = \frac{hc}{\lambda} \] Substitute the values:
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{500} = \frac{1.989 \times 10^{-25}}{500} = 3.978 \times 10^{-28} \, \text{J} \]

Step 2: Calculate the number of photons emitted per second:
The transmitter power \( P \) is the energy emitted per second.
So, number of photons emitted per second \( n \) is:
\[ n = \frac{P}{E} = \frac{10^{4}}{3.978 \times 10^{-28}} \approx 2.51 \times 10^{31} \]

Step 3: Order of magnitude:
\[ n \approx 10^{31} \]

Therefore, the number of photons emitted per second by the transmitter is of the order of:
\[ \boxed{10^{31}} \]