Question

The momentum of photon of frequency ${10}^{14}Hz$ will be

• A. $2.2\times {10}^{-26}Kgm/sec$
• B. $2.21\times {10}^{-28}Kgm/sec$
• C. ${10}^{-28}Kgm/sec$
• D. $0.21\times {10}^{-2}Kgm/sec$

Answer 1

The correct option is B $2.21\times {10}^{-28}Kgm/sec$

From de Broglie hypothesis, wavelength of photon , $\lambda =\frac{h}{p}$

or $\frac{c}{f}=\frac{h}{p}$ where $h=$ Planck's constant , $p=$ momentum , $f=$ frequency and $c=$ velocity of light or photon.

So, $p=\frac{hf}{c}=\frac{(6.62\times {10}^{-34})\times {10}^{14}}{(3\times {10}^8)}=2.21\times {10}^{-28}Kgm/sec$

So, the answer is option (B).