The work function of a certain metal is \( 3.31\times10^{-19}\,\mathrm{J} \). The maximum kinetic energy of the photoelectrons emitted by incident radiation of wavelength 5000 Å is (take \( h=6.63\times10^{-34}\,\mathrm{J\,s} \), \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.42 eV
The maximum kinetic energy of the photoelectrons is the photon energy \( h\nu=hc/\lambda \) minus the work function. This is Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \).
\( \lambda=5000\times10^{-10}\,\mathrm{m}=5\times10^{-7}\,\mathrm{m} \)
The energy of each photon is:
\( \begin{aligned} h\nu &= \frac{hc}{\lambda} \\ &= \frac{6.63\times10^{-34}\,\mathrm{J\,s}\times3\times10^{8}\,\mathrm{m\,s^{-1}}}{5\times10^{-7}\,\mathrm{m}} \\ &= 3.978\times10^{-19}\,\mathrm{J} \end{aligned} \)
The maximum kinetic energy is:
\( \begin{aligned} K_{\mathrm{max}} &= h\nu-\phi_0 \\ &= 3.978\times10^{-19}\,\mathrm{J}-3.31\times10^{-19}\,\mathrm{J} \\ &= 0.668\times10^{-19}\,\mathrm{J} \end{aligned} \)
In electron volts:
\( \begin{aligned} K_{\mathrm{max}} &= \frac{0.668\times10^{-19}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{J/eV}} \\ &\approx 0.42\,\mathrm{eV} \end{aligned} \)