Question

The photosensitive surface is receiving the light of wavelength $5000\overset{^\circ }{A}$ at the rate of $10^{- 8} \, J s^{- 1}.$ The number of photons received per second is ( $h=6.62\times 10^{- 34}Js,c=3\times 10^{8}ms^{- 1}$ )

• A. $2.5\times 10^{10}$
• B. $2.5\times 10^{11}$
• C. $2.5\times 10^{12}$
• D. $2.5\times 10^{9}$

Answer 1

Answer: (A)

Energy of photon
$ \, E=\frac{h c}{\lambda }$
Given, $ \, \, \, \lambda =5000 \, \text{Å} \, = \, 5\times 10^{- 7 \, }\text{m}$
$\therefore \, \, E=\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}}$
$= \, 3.96\times 10^{- 19} \, \text{J}$
Energy received per second = $10^{- 8} \, J \, s^{- 1}$
$\therefore $ Number of photon's received per second
= $\frac{E n e r g y \, r e c e i v e d \, p e r \, s e c o n d \, }{E n e r g y \, o f \, o n e \, p h o t o n \, }$
= $\frac{10^{- 8}}{3.96 \times 10^{- 19}}=2.5\times 10^{10}$