The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm,

Question:

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1 \mathrm{~mW}$ and wavelength of $1000 \mathrm{~nm}$,

in $0.1$ second is $x \times 10^{13}$. The value of $x$ is________ (Nearest integer)

$\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{c}=3.00 \times 10^{8} \mathrm{~ms}^{-1}\right)$

Solution:

Energy emitted in $0.1 \mathrm{sec}$.

$=0.1 \mathrm{sec} \cdot \times 10^{-3} \frac{\mathrm{J}}{\mathrm{s}}$

$=10^{-4} \mathrm{~J}$

If 'n' photons of $\lambda=1000 \mathrm{~nm}$ are emitted,

then $; 10^{-4}=\mathrm{n} \times \frac{\mathrm{hc}}{\lambda}$

$\Rightarrow 10^{-4}=\frac{\mathrm{n} \times 6.63 \times 10^{-34} \times 3 \times 10^{8}}{1000 \times 10^{-9}}$

$\Rightarrow \mathrm{n}=5.02 \times 10^{14}=50.2 \times 10^{13}$

$\Rightarrow 50$ (nearest integer)

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