When a piece of metal is illuminated by a monochromatics light of wavelength `gamma` then stopping potential is `3V_(s)`. When same surface is illuminated by light of wavelength `2gamma` , then stopping potential becomes `V_(s)` . The value of threshold wavelength for photoelectric emission will be

Options

A. `(4gamma)//3`
B. `6gamma`
C. `4gamma`
D. `8gamma`

Answer

C

Explanation

`hv=W_(0)+T_(e)`(max)
`(hc)/(lambda)=W_(0)+3V_(s)` ……. (i)
`(hc)/(2lambda)=W_(0)+V_(s)` ………. (ii)
(i) - (ii) `therefore(hc)/(lambda)-(hc)/(2lambda)=(hc)/(2lambda)=2V_(s)therefore(hc)/(lambda)=4V_(s)`.
`thereforeW_(0)=V_(s)therefore(hc)/(lambda_(0))=(hc)/(4lambda)rArrlambda_(0)=4lambda`.