In an experiment on the photoelectric effect, it was observed that for incident light of wavelength \( 1.98\times10^{-7}\,\mathrm{m} \), the stopping potential is 2.5 V. What is the work function \( \phi_0 \) of the metal? (Take \( hc=1.98\times10^{-25}\,\mathrm{J\,m} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3.75 eV
By Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \), and the stopping potential \( V_0 \) measures the maximum kinetic energy, \( K_{\mathrm{max}}=eV_0 \). So the work function is the photon energy minus \( eV_0 \).
The energy of each photon is:
\( \begin{aligned} h\nu &= \frac{hc}{\lambda} \\ &= \frac{1.98\times10^{-25}\,\mathrm{J\,m}}{1.98\times10^{-7}\,\mathrm{m}} \\ &= 1\times10^{-18}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} h\nu &= \frac{1\times10^{-18}}{1.6\times10^{-19}}\,\mathrm{eV} \\ &= 6.25\,\mathrm{eV} \end{aligned} \)
With \( V_0=2.5\,\mathrm{V} \), the maximum kinetic energy is \( K_{\mathrm{max}}=2.5\,\mathrm{eV} \). The work function is:
\( \begin{aligned} \phi_0 &= h\nu-K_{\mathrm{max}} \\ &= 6.25\,\mathrm{eV}-2.5\,\mathrm{eV} \\ &= 3.75\,\mathrm{eV} \end{aligned} \)