Question

If the radiation of frequency ν is incident on a photosensitive metal, the maximum kinetic energy of the photoelectrons ejected is E. If the frequency of incident radiation is halved, find out the maximum kinetic energy of the photoelectrons. (Given ν/2 is greater than the threshold frequency)

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

Answer 1

The correct option is D $E-\frac{h\nu }{2}$

Energy of the 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 respectively. When the frequency of incident radiation is halved i.e ${\nu }_{new}=\frac{\nu }{2}$ then consider new kinetic energy is $E_1$ using equation (1) an equation (2)
$\frac{h\nu }{2}=h{\nu }_0+E_1\Rightarrow \frac{h\nu }{2}=(h\nu -E)+E_1$
$E_1=E-\frac{h\nu }{2}$