When the light of frequency $2 v_0$ (where $v_0$ is the threshold frequency) is incident on a photosensitive material, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident raditation is increased to $3 v_{0}$, the maximum velocity of emitted electrons from the same material is $v_2$. The ratio of $v_1$ to $v_2$ will then be:

$1: \sqrt{2}$

The correct answer is Option (4) → $1: \sqrt{2}$

$K.E_{max}=hf-hf_0$ [Einstein photoelectric equation]

where,

f = frequency of light

$f_0$ = threshold frequency

and,

$K.E_{max}=\frac{1}{2}mv^2$

v, velocity of $e^-$ = maximum velocity

Case 1: for frequency $2f_0$

$K.E_{max}=h(2f_0-f_0)=hf_0$

and,

$hf_0=\frac{1}{2}mv_1^2$

$v_1=\sqrt{\frac{2hf_0}{m}}$

Case 2: for frequency $3f_0$

$K.E_{max}=h(3f_0-f_0)=2hf_0$

and,

$2hf_0=\frac{1}{2}mv_2^2$

$v_2=\sqrt{\frac{4hf_0}{m}}$

$∴\frac{v_1}{v_2}=\frac{\sqrt{\frac{2hf_0}{m}}}{\sqrt{\frac{4hf_0}{m}}}=\frac{1}{\sqrt{2}}$