When a metallic surface is illuminated with radiation of wavelength `lamda`, the stopping poential is V. If the same surface is illuminated with radiation of wavelength `2lamda`, the stopping potential is `V/4`. The threshold wavelength for the metallic surface is

Options

A. `5lamda`
B. `(5)/(2) lamda`
C. `3 lamda`
D. `4 lamda`

Answer

C

Explanation

In the first case, `eV = (hc)/(lamda) - W_(0)`…(1)
In the second case, `e(V)/(4) = (hc)/(2lamda) - W_(0)` …(2)
Substracting equation `(2) xx 4` from equation (1), we get,
`0 = - (hc)/(lamda) + 3W_(0) or, W_(0) = (hc)/(3 lamda)`
`:.` Threshold wavelength, `lamda_(0) = 3 lamda`