Question

A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is $4.3\times {10}^{14}Hz$. The velocity of ejected electron is $\times {10}^{5}m{s}^{-1}.$ (Nearest integer)
[Use : $h=6.63\times {10}^{-34}Js,m_e=9.0\times {10}^{-31}kg$]

Answer 1

$h\nu =h{\nu }_0+\frac{1}{2}{m}_e{v}^2$
$\frac{hc}{\lambda }=h{\nu }_0+\frac{1}{2}{m}_e{v}^2$
Here,
${\nu }_0=\text{threshold frequency of the metal}$
$h\text{is planck's constant}$
$m_e\text{is mass of electron}$
$v\text{is velocity of ejected electron}$
Putting values:

$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-9}}=6.63\times {10}^{-34}\times 4.3\times {10}^{14}+\frac{1}{2}\times 9\times {10}^{-31}\times v^2$

$v=5\times {10}^{5}m{s}^{-1}$