JEE Physics 2026 Question with Solution

Given: Work function $\phi = 110 \times 10^{-20} \, \text{J}$ and Planck's constant $h = 6.63 \times 10^{-34} \, \text{J\cdot s}$. The photoelectrons have zero kinetic energy.

Find: The angular frequency of the incident light.

According to Einstein’s photoelectric equation,

$$h\nu = \phi + K_{\max}$$

Since the photoelectrons have zero kinetic energy,

$$K_{\max} = 0$$

Therefore,

$$h\nu = \phi$$

Now substitute the given values:

$$\nu = \frac{\phi}{h} = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}} = 1.66 \times 10^{15} \; \text{Hz}$$

Angular frequency is

$$\omega = 2\pi\nu$$

So,

$$\omega = 2\pi \times 1.66 \times 10^{15} \approx 1.04 \times 10^{16} \; \text{rad/s}$$

Therefore, the angular frequency is $1.04 \times 10^{16} \, \text{rad/s}$ and the correct option is D.

Given: The emitted photoelectrons have zero kinetic energy.

Find: The angular frequency of the incident radiation.

When the kinetic energy of emitted photoelectrons is zero, the incident frequency equals the threshold frequency. Hence,

$$\nu = \nu_0$$

and the threshold condition is

$$h\nu_0 = \phi$$

Thus,

$$\nu_0 = \frac{\phi}{h}$$

Substituting,

$$\nu_0 = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}} = 1.66 \times 10^{15} \; \text{Hz}$$

Now convert ordinary frequency to angular frequency using

$$\omega = 2\pi\nu$$

Hence,

$$\omega = 2\pi \times 1.66 \times 10^{15} \approx 1.04 \times 10^{16} \; \text{rad/s}$$

Therefore, the correct option is D.