Q51 / 75 · JEE Main 2026 (22 January Shift 2) · Physics
Light is incident on a metallic plate having work function $110 \times 10^{-20} \mathrm{~J}$. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is $\_\_\_\_$ rad/s. $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}\right)$.
- A$1.04 \times 10^{13}$
- Correct answerB$1.04 \times 10^{16}$
- C$1.66 \times 10^{16}$
- D$1.66 \times 10^{15}$
Solution
From Einstein's photoelectric equation, $h\nu = W + KE$.
With zero kinetic energy, $h\nu = W$.
The frequency is $\nu = \frac{W}{h} = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}} = 16.59 \times 10^{14}$ Hz.
The angular frequency is $\omega = 2\pi\nu = 2\pi \times 16.59 \times 10^{14}$
$= 104.25 \times 10^{14} = 1.04 \times 10^{16}$ rad/s.
With zero kinetic energy, $h\nu = W$.
The frequency is $\nu = \frac{W}{h} = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}} = 16.59 \times 10^{14}$ Hz.
The angular frequency is $\omega = 2\pi\nu = 2\pi \times 16.59 \times 10^{14}$
$= 104.25 \times 10^{14} = 1.04 \times 10^{16}$ rad/s.
Hence, the correct answer is option: 'B'.