Question

A monochromatic source of light emits photons of frequency 6 × 10¹⁴ Hz. The power emitted by the source is 8 × 10^-3 W. Calculate the number of photons emitted per second (Take h = 6.63 × 10^-34 J-s)
1. 6 × 10¹⁴
2. 4 × 10¹⁵
3. 2 × 10¹⁶
4. 1 × 10¹⁷

Answer 1

Correct Answer - Option 3 : 2 × 10¹⁶

Concept:

Energy of photon is given by:

E = hν

where h is Planck's constant = 6.63 × 10-34 J-s, ν = frequency

Number of photons emitted per second (N) from a source is given by:

\(N=\frac{P}{E}\)

where P is power of source, E is energy of photon

Calculation:

Given:

ν = 6 × 1014 Hz, P = 8 × 10-3 W

Energy of photon is:

E = hν

E = 6.63 × 10-34 × 6 × 10¹⁴ J = 39.8 × 10^-20 J

Number of photons emitted per second (N) from a source is:

\(N=\frac{P}{E}\)

\(N=\frac{8\times10^{-3}}{39.8\times10^{-20}}\)= 2 × 1016