When a metallic surface is illuminated with light of photon energy 6 eV, the stopping potential is \( 5V_0 \). When the same surface is illuminated with light of photon energy 2 eV (three times the wavelength), the stopping potential is \( V_0 \). The work function of the metallic surface is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 eV
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons is \( K_{\mathrm{max}}=E-\phi_0 \), and the stopping potential gives \( K_{\mathrm{max}}=eV_0 \). So for each light, \( eV=E-\phi_0 \).
For the two lights:
\( 5eV_0=6\,\mathrm{eV}-\phi_0 \)
\( eV_0=2\,\mathrm{eV}-\phi_0 \)
Subtracting the second equation from the first:
\( 4eV_0=4\,\mathrm{eV} \)
\( \Rightarrow\; eV_0=1\,\mathrm{eV} \)
Putting this in the second equation:
\( \begin{aligned} \phi_0 &= 2\,\mathrm{eV}-eV_0 \\ &= 2\,\mathrm{eV}-1\,\mathrm{eV} \\ &= 1\,\mathrm{eV} \end{aligned} \)