Question

The work function of sodium metal is $4.41\times 10^{- 19}J$ . If photons of wavelength $300\,nm$ are incident on the metal, the kinetic energy of the ejected electrons will be $\left(h = 6 . 63 \times 10^{- 34} J s ; c = 3 \times 10^{8} m / s\right)$ _______ $\times 10^{- 21}J$ .

Answer 1

$\phi=4.41\times 10^{- 19}J$
$\lambda =300\,nm$
$KE_{\max}=\frac{hc}{\lambda }-\phi$
$=\frac{6 . 63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}}-4.41\times 10^{- 19}$
$=6.63\times 10^{- 19}-4.41\times 10^{- 19}$
$=222\times 10^{- 21}J$