Question

If a light with frequency $4\times {10}^{16}Hz$ emitted photoelectrons with double the maximum kinetic energy as are emitted by the light of frequency $2.5\times {10}^{16}Hz$ from the same metal surface, if the threshold frequency $\left(v_0\right)$ of the metal is $x\times {10}^{16}Hz.$ Then the value of x:

Answer 1

According to Einstein equation,
$hv=h{v}_o+K{E}_{max}$
Where, ${\nu }_o=\text{threshold frequency}$
For frequency $2.5\times {10}^{16}Hz,$
$\therefore h(2.5\times {10}^{16})=h{v}_0+K{E}_{max}.......\left(1\right)$
For frequency $4\times {10}^{16}Hz,$
$h(4\times {10}^{16})=h{v}_0+2K{E}_{max}.......\left(2\right)$
Multiply eqn (1) by 2 and subtract eqn (2) from (1).
$\therefore 2h{v}_0-h{v}_0=h(5.0\times {10}^{16}-4\times {10}^{16})$
$\therefore h{v}_o=h(1\times {10}^{16})$
$\therefore$ threshold frequency, $v_o=1\times {10}^{16}Hz$