A certain metallic surface is illuminated with monochromatic light of wavelength `lambda`.The stopping potentail for photelectric current for this light is `3V_(0)`.If 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. `4lambda`
B. `(lambda)/(4)`
C. `(lambda)/(6)`
D. 6`lambda`

Answer

A

Explanation

According to equation of Einstein,
`eV_(s)=E-phi`
`therefore V_(s)=(hc)/(lambda_(e))-(hc)/(lambda_(0)e)`
`therefore 3V_(0)=(hc)/(lamdbae)-(hc)/(lambda_(o)e)`…..(1) and
`therefore V_(o)=(hc)/(2lambdae)-(hc)/(lambda_(0)e)`…….(2)
By multiplying eq. (1) with (2) and then substracting from (1),
`3V_(0)=(hc)/(lambdae)-(hc)/(lambda_(0)e)`
`3V_(0)=(3hc)/(2lambdae)-(3hc)/(lambda_(0)e)`
`(--+)/(0=-(hc)/(2lambdae)+(2hc)/(lambda_(0)e))`
`0=-(hc)/(2lambdae)+(2hc)/(lambda_(0))`
`therefore (2)/(lambda_(0))=(1)/(2lambda)`
`therefore lambda_(0)=4lambda`