Question

A certain metal was irradiated with light of frequency $3.2\times {10}^{16}se{c}^{-1}$. The photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with a light of frequency $2\times {10}^{16}se{c}^{-1}$. Calculate the threshold frequency of the metal.

Answer 1

according to question K.E of photoelectrons when metal plate treated with light of frequency $3.2\times {10}^{16}se{c}^{-1}$ $=h(3.2\times {10}^{16}-2\times {10}^{16})=h(1.2\times {10}^{16})$

$v_0=$ threshold frequency

$h{v}_0=h(3.2\times {10}^{16})-h(1.2\times {10}^{16})$

$v_0={10}^{16}$ $se{c}^{-1}$