When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8V.

Question

When a metal surface is illuminated by light of wavelength \(\lambda\), the stopping potential is 8V. When the same surface is illuminated by light of wavelength \(3\lambda\), stopping potential is 2V. The threshold wavelength for this surface is : 

(1) \(5\lambda\) 

(2) \(3\lambda\)

(3) \(9\lambda\)

(4) \(4.5 \lambda\)

Answer 1

Correct option is (3) \(9\lambda\)

\(E = \phi + K_{max}\)

\(\phi = \frac {hc}{\lambda _ 0}\)

\(K_{max} = eV_0\)

\(8e = \frac {hc}{\lambda} - \frac {hc}{\lambda _0} \,\,...(i)\)

\(2e = \frac {hc}{3\lambda} - \frac {hc}{\lambda _0}\,\,...(ii)\)

on solving (i) & (ii)

\(\lambda _0 = 9\lambda\)