QJEEM17Ph35aa: JEE Main 2017 Physics - Class 12 JEEM17Ph35aa Solution: The maximum velocity of the photoelectrons emitted from the…

The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

$KE = {1 \over 2}m{v^2} = hn - \phi$

where h is Planck's constant and $\phi$ is the work function. Therefore, the kinetic energy of the incident light is

${E_1} = hn - \phi$ ..... (1)

When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

${1 \over 2}mv_1^2 = 3hn - \phi$

$\Rightarrow {E_2} = 3hn - \phi$ ..... (2)

Substituting $hn = {E_1} + \phi$ [from Eq. (1)] in Eq. (2), we get

${E_2} = 3({E_1} + \phi ) - \phi$

$\Rightarrow {E_2} = 3{E_1} + 2\phi$

$\Rightarrow {1 \over 2}mv_1^2 = 3 \times {1 \over 2}m{v^2} + 2\phi$

$\Rightarrow v_1^2 = 3{v^2} + 2\phi \times {2 \over m}$

$\Rightarrow v_1^2 = 3{v^2} + {{4\phi } \over m}$

Thus, the velocity is more than $\sqrt 3 v$ .