Question

The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\text{nm},$ when it delivers the power of $3.3\times {10}^{–3}\text{watt}$ will be $(h=6.6\times {10}^{–34}\text{Js})$

• A. ${10}^{15}$
• B. ${10}^{18}$
• C. ${10}^{17}$
• D. ${10}^{16}$

Answer 1

The correct option is D ${10}^{16}$

The power of a source is given as,

$P=\frac{E}{t}=\frac{n}{t}\left(\frac{hc}{\lambda }\right)$

$\Rightarrow \frac{n}{t}=\frac{P}{\left(\frac{hc}{\lambda }\right)}$

(Here $\frac{n}{t}$ is number of photons emitted per second)

$\Rightarrow \frac{n}{t}=\frac{3.3\times {10}^{-3}\times 600\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}={10}^{16}\text{photons/sec}.$

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