When light of wavelength `lambda` is incident on a metal surface, stopping potential is found to be x. When light of wavelength n`lambda` is incident on the same metal surface, stopping potential is found to be `(x)/(n+1)` . Find the threshold wavelength of the metal.

Options

A. `n/lambda`
B. `(n^2)/(lambda)`
C. `n lambda`
D. `n^2 lambda`

Answer

D

Explanation

Let `lambda_0` is thee threshold wavelegth , the work function is `phi = (hc)/(lambda_0)`
Now, by photoelectric equation `eV= (hc)/(lambda)-(hc)/(lambda_0)`
`ex = (hc)/(lambda)-(hc)/(lambda_0)`
`(ex)/(n+1) = (hc)/(n lambda)- (hc)/(lambda_0)`
From (i) and (ii), `(hc)/(lambda)-(hc)/(lambda_0) = (n+1) (hc)/(n lambda)-(n+1) (hc)/(lambda_0)`
or `(nhc)/(lambda_0) = (hc)/(n lambda) rArr lambda_0 = n^2 lambda`