By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from `4xx10^(15)Hz` to `8xx10^(15)Hz`? Given, `h=6.6xx10^(-34)Js`, `e=1.6xx10^(-19)C and c=3xx10^8ms^-1`.

Options

A. 6 V
B. 8 V
C. 12 V
D. 16 V

Answer

D

Explanation

According to Einstein's photoelectric equation, the maximum K.E. of the electrons is
`K_("max")=eV_(0)=hv-W_(0)`
`:.eV_(1)=hv_(1)-W_(0)`
and `eV_(2)=hv_(2)-W_(0)`
`e(V_(2)-V_(1))=h(v_(2)-v_(1))`
`:.V_(2)-V_(1)=h/e(v_(2)-v_(1))`
`=(6.4xx10^(-34)xx(8-4)10^(15))/(1.6xx10^(-19))`
`=16V`