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

To find the number of photons emitted per second, we use the formula for energy of a photon, \(E = h \cdot f\), where \(h\) is Planck's constant \((6.626 \times 10^{-34} \, \text{Js})\) and \(f\) is the frequency. We can rearrange this to find the number of photons: 

1. The energy of one photon is calculated as:

\(E = 6.626 \times 10^{-34} \, \text{Js} \times 5.0 \times 10^{14} \, \text{Hz} = 3.313 \times 10^{-19} \, \text{J}\)

1. The power emitted by the source is given as 3.31 mW (\(3.31 \times 10^{-3} \, \text{W}\)). Power is energy per unit time, so to find the number of photons emitted per second, we use:

\(n = \dfrac{\text{Power}}{\text{Energy per photon}} = \dfrac{3.31 \times 10^{-3} \, \text{W}}{3.313 \times 10^{-19} \, \text{J/photon}}\)

1. Solving the above expression gives:

\(n = \dfrac{3.31 \times 10^{-3}}{3.313 \times 10^{-19}} \approx 1.0 \times 10^{16}\)

Therefore, the number of photons emitted per second is approximately \(10^{16}\). The correct option is: \(10^{16}\)