Question

An incident radiation has wavelength $900\text{nm}$ associated with it. Find the power rating of the source, if it is working at $30\text{\%}$ efficiency and $4\times {10}^{25}$ photons are crossing unit cross-section per second.

• A. $40.45\text{MW}$
• B. $29.45\text{MW}$
• C. $46.45\text{MW}$
• D. $20.45\text{MW}$

Answer 1

The correct option is B $29.45\text{MW}$

Energy of each photon $=\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{900\times {10}^{-9}}$

$\therefore E=2.21\times {10}^{-19}\text{J}$

$\text{Total power of incident radiation}=$
$\text{no. of photons incident per second}\times \text{Energy associated with each photon}$

$=4\times {10}^{25}\times 2.21\times {10}^{19}$
$=8.84\times {10}^6\text{W}$
$\text{Power rating}=P=\frac{\text{Power}}{\text{Efficiency}}$
$=\frac{8.84\times {10}^6}{30\times {10}^{-2}}$
$=29.46\times {10}^6\text{W}=29.46\text{MW}$

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

$\begin{array}{c}\text{Why this question?}\\ \text{This question deals with how each photon contributes to the power of incident radiation. (Typical formula based question.)}\end{array}$