When a certain metallic surface is illuminated with mono chromatic light of wavelength `lambda`, the stopping potential for photoelectric current is `3V_(0)`. When the same surface is illuminated with light of wavelength `2lambda` the stopping potential is `V_(0)`. The threshold wavelength for this surface for photoelectric effect is.

Options

A. `(pi)/(4)`
B. `(lambda)/(6)`
C. `6lambda`
D. `4lambda`

Answer

4

Explanation

According to Einstein.s photoelectric equation
`V_(s) = (hc)/(e) ((1)/(lambda)-(1)/(lambda_(0)))`
For the first case
`3V_(0) = (hc)/(e) ((1)/(lambda)-(1)/(lambda_(0)))`
For the second case
`V_(0)=(hc)/(e) ((1)/(2lambda)-(1)/(lambda_(0)))`
Divide eqn. (i) by (ii), we get
`lambda_(0) = 4 lambda`