In an experiment of photoelectric emission for incident light of `4000Å`, the stopping potential is 2V. If the wavelength of incident light is made `3000Å`, then stopping potential will be :

Options

A. less than 2 volt
B. More than 2 volt
C. 2 volt
D. zero.

Answer

B

Explanation

`(hc)/(lambda_(1))-(phi_(0))/(e)=V_(0_(1))" "...(1)`
`(hc)/(lambda_(2))-(phi_(0))/(e)=V_(0_(2)) " "...(2)`
From eqs (2)-(1)
`rArr V_(0_(2))-V_(0_(1))=(hc)/(elambda_(2))-(hc)/(elambda_(1))" "...(3)`
`lambda_(1)=400xx10^(-10)m`
`lambda_(2)=3000xx10^(-10)m" "...(4)`
`V_(0_(1))=2` volt
`V_(0_(2))=?`
From eqs (3) and (4),
`V_(0_(2))2+(6.62xx10^(-34)xx3xx10^(8))/(1*6xx10^(-19)) xx[(1)/(3xx10^(-7))-(1)/(4xx10^(-7))]`
`V_(0_(2))=2+1=3` volts.