The threshold frequency of a metal is f_0 . When the light of frequency 2 f_0 is incident on the metal plate, the maximum velocity of photoelectrons is v_1 . When the frequency of incident radiation is increased to 5 f_0 , the maximum velocity of photoelectrons emitted is v_2 . The ratio of v_1 to v_2 is :

Dual Nature of Radiation PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q61 with Solution

View written solutionFree

Correct answer: A

For a metal with threshold frequency $f_0$ , the work function is

\phi = h f_0

The maximum kinetic energy of emitted photoelectrons is

K_{\max} = h f - \phi = h(f-f_0)

Also,

K_{\max} = \frac{1}{2} m v_{\max}^2

So,

\frac{1}{2} m v^2 = h(f-f_0)

Then,

\frac{1}{2} m v_1^2 = h(2f_0 - f_0) = h f_0

So,

v_1^2 = \frac{2 h f_0}{m}

Then,

\frac{1}{2} m v_2^2 = h(5f_0 - f_0) = 4 h f_0

So,

v_2^2 = \frac{8 h f_0}{m}

\frac{v_1^2}{v_2^2} = \frac{\frac{2 h f_0}{m}}{\frac{8 h f_0}{m}} = \frac{1}{4}

Therefore,

\frac{v_1}{v_2} = \frac{1}{2}

So the correct option is A.