Question

If the maximum kinetic energy of the photoelectrons is $E$ when a radiation of frequency $\nu$ is incident on a photosensitive metal, then the maximum kinetic energy of the photoelectrons when the frequency of the incident radiation is doubled is:

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

Answer 1

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

We know,
$\text{Energy of incident radiation = work function+KE}$
Given, maximum kinetic energy = $E$
So,
$h\nu =h{\nu }_o+E$ ...(1)
Here, $\nu$ and ${\nu }_o$ are the incident and the threshold frequencies respectively.

Let's assume that when the frequency of the incident radiation is doubled the new maximum kinetic energy = $E^{\prime }$.

Therefore,
$2h\nu =h{\nu }_o+E^{\prime }$
$\Rightarrow 2h\nu =(h\nu -E)+E^{\prime }\text{[From equation 1]}$
$\Rightarrow E^{\prime }=E+h\nu$