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 \[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{500} \] \[ E = \frac{1.9878 \times 10^{-25}}{500} \] \[ E = 3.9756 \times 10^{-28} { J} \] 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 \( 10^{31} \).