When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :

Options

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

Answer

D

Explanation

Accoding to einstein.s photoelectric effect
`eV =(hc)/(lamda) - (hc)/(lamda_0)` …(i)
`eV//4 = (hc)/(2lamda) - (hc)/(lamda_0) ` …(ii)
Dividing equation (i) by (ii) by
`rArr 4 = (1/lamda -1lamda_0)/((1)/(2lamda) - (1)/(lamda_0) ) `on solving we get , `lamda_0 = 3 lamda `