A) \[2\,\,V\]
B) less than\[2\,\,V\]
C) more than\[2\,\,V\]
D) zero
Correct Answer: C
Solution :
When stopping potential is\[V\], then \[E=eV-\phi \] or \[\frac{hc}{\lambda }=eV-\phi \] or \[eV=\frac{hc}{\lambda }+\phi \] where\[e,\,\,h\]and\[c\]are constants \[\therefore \] \[V\propto \frac{1}{\lambda }\] Hence, if wavelength decreases, stopping potential will increase. \[\therefore \] \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{4000}{3000}\] or \[{{V}_{2}}=\frac{4}{3}\times 2v=\frac{8}{3}v>2v\]
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