Question

The photon energy in units of eV for electromagnetic waves of wavelength $2cm$ is:

• A. $2.5{10}^{-19}$
• B. $5.2{10}^{-16}$
• C. $3.2{10}^{-16}$
• D. $6.2{10}^{-5}$

Answer 1

Answer: (D)

$6.2{10}^{-5}$

Given,

$\lambda =2cm=0.02m$

$h=6.62\times {10}^{-34}kg{m}^2/s$

$c=3\times {10}^{8}m/s$

Photon energy, $E=\frac{hc}{\lambda }$

$E=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{0.02}$

$E=993\times {10}^{-26}J$

Photon energy in unit of $eV$,

$E=\frac{993\times {10}^{-26}}{1.6\times {10}^{-19}}eV$

$E=6.20\times {10}^{-5}eV$