Question

Calculate the maximum kinetic energy of photoelectrons emitted when a light of frequency $2\times {10}^{16}Hz$ is irradiated on a metal surface with threshold frequency $v_0$ equal to $8.68\times {10}^{15}Hz$

• A. $7.50\times {10}^{-8}J$
• B. $8.38\times {10}^{-18}J$
• C. $7.50\times {10}^{-18}J$
• D. $8.38\times {10}^{-8}J$

Answer 1

The correct option is C $7.50\times {10}^{-18}J$

$h\nu =h{\nu }_0+KE$
Threshold frequency $\left({\nu }_0\right)=8.68\times {10}^{15}Hz/s$
Frequency of light $\nu )=2\times {10}^{16}Hz$
$K.E=h(\nu -{\nu }_0)$
$=6.626\times {10}^{-34}(2\times {10}^{16}-8.68\times {10}^{15})$
$K.E=7.5\times {10}^{-18}J$
Maximum kinetic energy of photoelectrons = $7.5\times {10}^{-18}J$