Question

Radiation of frequency ν is incident on photosensitive metal. Maximum kinetic energy of the photoelectrons is E. If the frequency of incident radiation is doubled, find out the maximum kinetic energy of photoelectrons.

• A. 2E
• B. $\frac{E}{2}$
• C. $E+h\nu$
• D. $E-h\nu$

Answer 1

The correct option is C $E+h\nu$

Energy incident Radiation = work function + kinetic energy ... (1)
Equation (1) becomes
$h\nu =h{\nu }_0+E\dots .\left(2\right)$
Where $\nu$ and ${\nu }_0$ are incident and threshold frequencies. When frequency of incident radiation is doubled i.e $\nu =2\nu$ then consider new kinetic energy is $E_1$ using equation (1) an equation (2)
$h\left(2\nu \right)=h{\nu }_0+E_1\Rightarrow 2h\nu =(h\nu -E)+E_1$
$E_1=-(h\nu -E)+2h\nu$
$E_1=E+h\nu$