When a certain metallic surface is illuminated with monochromatic light wavelength \(\lambda\), the stopping potential for photoelectric current is \(4 V_0\). When the same surface is illuminated with light of wavelength \(3 \lambda\), the stopping potential is \(\mathrm{V}_0\). The threshold wavelength for this surface for photoelectric effect is

Using Einstein's photoelectric equation, \(\mathrm{h} v=\phi_0+\mathrm{KE}_{\text {max }}\)
\(\therefore \quad \frac{\mathrm{hc}}{\lambda}=\phi_0+\mathrm{e}\left(4 \mathrm{~V}_0\right)\)
....(i) \(\left(\because \mathrm{KE}_{\max }=\mathrm{eV}_{\mathrm{s}}\right)\)
Also, \(\frac{\mathrm{hc}}{3 \lambda}=\phi_0+\mathrm{eV}_0\)...(ii)
Subtracting equation (i) from \(4 \times\) equation (ii) we get,
\(\left(\frac{4}{3}-1\right) \frac{\mathrm{hc}}{\lambda}=4 \phi_0-\phi_0\) or \(\phi_0=\frac{\mathrm{hc}}{9 \lambda}\)
But \(\phi_0=\frac{\mathrm{hc}}{\lambda_0}\), where \(\lambda_0\) is the threshold wavelength, hence \(\lambda_0=9 \lambda\).
Hence, option (A) is correct.