Question

A source produces monochromatic light of frequency \( 5.0 \times 10^{14} \, \text{Hz} \) and the power emitted is 3.31 mW. The number of photons emitted per second by the source, on an average, is:

• A. \( 10^{16} \)
• B. \( 10^{24} \)
• C. \( 10^{10} \)
• D. \( 10^{20} \)

Hint:

To find the number of photons emitted per second, use the formula \( \text{Number of photons} = \frac{P}{h \nu} \), where \( P \) is power and \( \nu \) is the frequency of light.

Answer 1

The Correct Option is A

The energy of a single photon is given by the formula: \[ E = h \nu \] where \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \, \text{J s} \)) and \( \nu \) is the frequency of the light. Given: - Frequency \( \nu = 5.0 \times 10^{14} \, \text{Hz} \), - Power \( P = 3.31 \, \text{mW} = 3.31 \times 10^{-3} \, \text{W} \). The total number of photons emitted per second is given by: \[ \text{Number of photons} = \frac{P}{E} \] Substitute \( E = h \nu \) into this equation: \[ \text{Number of photons} = \frac{P}{h \nu} = \frac{3.31 \times 10^{-3}}{6.626 \times 10^{-34} \times 5.0 \times 10^{14}} = 10^{16} \] Thus, the correct answer is (A).