Question

A laser beam of frequency \( 3.0 \times 10^{14} \) Hz produces average power of 9 mW. Find (i) the energy of a photon of the beam, and (ii) the number of photons emitted per second on an average by the source.

Hint:

Use \( E = h\nu \) for single photon energy, and divide power by energy per photon to get the photon emission rate. Convert mW to W when using SI units.

Answer 1

(i) Energy of a photon:
The energy of a photon is given by the formula: \[ E = h \nu \] Where: \( h = 6.63 \times 10^{-34}\ \text{J·s} \) (Planck's constant) \( \nu = 3.0 \times 10^{14}\ \text{Hz} \) \[ E = 6.63 \times 10^{-34} \times 3.0 \times 10^{14} = 1.989 \times 10^{-19}\ \text{J} \] (ii) Number of photons emitted per second:
Power is the energy emitted per second. Given: \[ P = 9\ \text{mW} = 9 \times 10^{-3}\ \text{W} \] \[ \text{Number of photons per second} = \frac{P}{E} = \frac{9 \times 10^{-3}}{1.989 \times 10^{-19}} \approx 4.52 \times 10^{16} \] So, - Energy per photon = \( 1.989 \times 10^{-19} \) J - Number of photons per second = \( 4.52 \times 10^{16} \)