Question

The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency $2\times {10}^{14}{s}^{-1}$ is $6.63\times {10}^{-20}J$. Find the threshold frequency of the metal.

Answer 1

Absorbed energy =Threshold energy + Kinetic energy of photoelectrons

$h\vartheta =h{\vartheta }_o+KE$

$h{\vartheta }_0=h{\vartheta }^{}-KE$

$6.626\times 10^{-34}\times {\vartheta }_o=6.626\times 10^{-34}\times 2\times 10^{-14}-6.63\times 10^{-20}$

$v_o=\frac{1.3252\times 10^{-19}-6.63\times 10^{-20}}{6.626\times 10^{-34}}$

$v_o=9.99\times 10^{13}$

$=10^{14}{s}^{-1}$