Question

The threshold frequency for a photosenstive metal is $7\times {10}^{13}Hz$. If light of frequency ${10}^{14}Hz$ is incident on this metal, then maximum kinetic energy of emitted electron is?

• A. $4\times {10}^{-20}J$
• B. $8\times {10}^{-20}J$
• C. $2\times {10}^{-20}J$
• D. ${10}^{-20}J$

Answer 1

Answer: (C)

$2\times {10}^{-20}J$

We know,

$h(\nu -{\nu }_0)K{E}_{max}$

Given,

${\nu }_0=7\times {10}^{1}3$

$\nu ={10}^{1}4$

$K{E}_{max}=6.632\times 3\times {10}^{-34+13}=19.896\times {10}^{-21}=1.9896\times {10}^{-20}$

$K{E}_{max}=2\times {10}^{-20}$

Option $\text{C}$ is the correct answer