The work function of sodium metal is $4.41\times {10}^{-19}J$. If photons of wavelength $300nm$ are incident on the metal, the kinetic energy of the ejected electrons will be ($h=6.63\times {10}^{-34}Js;c=3\times {10}^{8}m/s$) _____ $\times {10}^{-21}J.$

Step 1:

Work function of sodium metal $=4.41\times {10}^{-19}J$
$\lambda ,$ wavelength of incident light = 300 nm
$=3\times {10}^{-7}m$

Step 2:
According to photoelectric effect,
$\frac{hc}{\lambda }=w+KE$
$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{3\times {10}^{-7}}=4.41\times {10}^{-19}+KE$

$6.63\times {10}^{-19}=4.41\times {10}^{-19}+KE$

$KE=2.22\times {10}^{-19}J=222\times {10}^{-21}J$