Question:
The work function of sodium metal is $4.41 \times 10^{-19} \mathrm{~J}$. If the photons of wavelength $300 \mathrm{~nm}$ are incident on the metal,
the kinetic energy of the ejected electrons will be
$\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js} ; \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
______$\times 10^{-21} \mathrm{~J}$.
Solution:
$E=W+K \cdot E_{\max }$
$\mathrm{K} \cdot \mathrm{E}_{\max }=\mathrm{E}-\mathrm{W}$
$=\frac{\mathrm{hc}}{\lambda}-4.41 \times 10^{-19}$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}}-4.41 \times 10^{-19}$
$=2.22 \times 10^{-19} \mathrm{~J}$
$=222 \times 10^{-21} \mathrm{~J}$