Question

A monochromic source of light operating at $60\text{W}$ emits $4\times {10}^{20}$ photons per second. Assuming the efficiency $\eta =50\text{\%}$, find the frequency of the photon.

• A. $1.133\times {10}^{12}{\text{s}}^{-1}$
• B. $1.133\times {10}^9{\text{s}}^{-1}$
• C. $4.133\times {10}^{14}{\text{s}}^{-1}$
• D. $1.133\times {10}^{14}{\text{s}}^{-1}$

Answer 1

The correct option is D $1.133\times {10}^{14}{\text{s}}^{-1}$

Given that,

Number of photon per second $n=4\times {10}^{20}{\text{s}}^{-1}$

Total energy of the source of light per second $E=60\text{W}\times 50\text{\%}$

$\Rightarrow E=30{\text{Js}}^{-1}$

Energy of each photon,

$E_1=\frac{30}{4\times {10}^{20}}=\frac{3}{4}\times {10}^{-19}\text{J}$

Also, energy of a photon, $E_1=hf$

$\Rightarrow f=\frac{E_1}{h}=\frac{3\times {10}^{-19}}{4\times 6.62\times {10}^{-34}}$

$\Rightarrow f=1.133\times {10}^{14}{\text{s}}^{-1}$

Hence, option $\left(D\right)$ is the correct answer.