Question

Photons of wavelength 662 nm are incident normally on a perfectly reflecting surface. If they exert a force of 1 N on the surface, then the number of photons incident per second is $(h=6.62\times {10}^{-34}Nm-s)$

• A. $5\times {10}^{26}$
• B. $2.5\times {10}^{26}$
• C. $1.25$
• D. $1\times {10}^{27}$

Answer 1

The correct option is A $5\times {10}^{26}$

Given,

$\lambda =662nm$

$h=6.62\times {10}^{-34}Nms$

$\Delta t=1sec$

$F=1N$

Lets consider $n$ is the number of photon

Change in momentum according to question,

$\Delta P=\frac{2h}{\lambda }$

From newton 2nd law,

$F=\frac{\Delta P}{\Delta t}$

$1=n\frac{2h/\lambda }{1}$

$n=\frac{\lambda }{2h}$

$n=\frac{662\times {10}^{-9}}{2\times 6.625\times {10}^{-34}}=5\times {10}^{26}$

The correct option is A.