JEE Physics 2023 Question with Solution

Given: The threshold frequency is $f_0$. For incident frequency $2f_0$, the maximum photoelectron velocity is $v_1$. For incident frequency $5f_0$, the maximum photoelectron velocity is $v_2$.

Find: The ratio $\frac{v_1}{v_2}$.

Using Einstein’s photoelectric equation,

$$K_{\text{max}} = hf - hf_0$$

and the kinetic energy relation,

$$K_{\text{max}} = \frac{1}{2}mv^2$$

For $f = 2f_0$,

$$\frac{1}{2}mv_1^2 = h(2f_0) - hf_0 = hf_0$$

For $f = 5f_0$,

$$\frac{1}{2}mv_2^2 = h(5f_0) - hf_0 = 4hf_0$$

Dividing the two equations,

$$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{hf_0}{4hf_0}$$ $$\frac{v_1^2}{v_2^2} = \frac{1}{4}$$

Therefore,

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

So the correct value is $\frac{1}{2}$. However, the solution marks Option D, while the listed options show $\frac{1}{2}$ as Option A. This is a source discrepancy.

Given: Maximum kinetic energy of photoelectrons depends on incident frequency through Einstein’s equation.

Find: How the ratio of velocities follows from the ratio of kinetic energies.

Since

$$\frac{1}{2}mv^2 \propto K_{\text{max}}$$

we get

$$v^2 \propto hf - hf_0$$

For the first case,

$$v_1^2 \propto 2f_0 - f_0 = f_0$$

For the second case,

$$v_2^2 \propto 5f_0 - f_0 = 4f_0$$

Hence,

$$\frac{v_1^2}{v_2^2} = \frac{1}{4}$$

and taking the positive square root,

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

Therefore, the defensible correct option from the listed choices is A.