Question

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is $x\times {10}^{13}.$ The value of x is . (Nearest integer)
($h=6.63\times {10}^{-34}Js,c=3.00\times {10}^{8}m{s}^{-1}$)

Answer 1

Power $=1mW$
$={10}^{-3}J$ in 1 sec.
$={10}^{-4}J$ in 0.1 sec.
$\therefore \text{Energy}=\frac{nhc}{\lambda }$
${10}^{-4}=\frac{n\times 6.63\times {10}^{-34}\times 3\times {10}^8}{1000\times {10}^{-9}}$
$n\left(numberofphotons\right)=50.2\times {10}^{13}$

Since

$n=x\times {10}^{13}$

$Hencevalueofx=50$

$\therefore x=50$