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 $\left({\nu }_o\right)$ 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 $7.50\times {10}^{-18}J$

We know,
$h\nu =h{\nu }_o+KE$
Given,
$\text{Threshold frequency}\left({\nu }_o\right)=8.68\times {10}^{15}Hz$
$\text{Frequency of light}\left(\nu \right)=2\times {10}^{16}Hz$

$\therefore$ KE = $h(\nu -{\nu }_o)$
$=6.626\times {10}^{-34}(2\times {10}^{16}-8.68\times {10}^{15})$
$=7.5\times {10}^{-18}J$