The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 α particles is \ \ \ \ kHz .[0pt] [mass of He nucleus = 4.002 amu, 1 amu =1.66 × 10 -27 kg , h =6.6 × 10 -34 J . s and c = 3 × 10 8 m / s ]

Physics · Nuclear Physics

JEE Main 2026 — 22 January, Morning Shift — Question 34

The minimum frequency of photon required to break a particle of mass 15.348 amu into $4 \alpha$ particles is $\_\_\_\_$ kHz .[0pt] [mass of He nucleus = 4.002 amu, $1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and $\mathrm{c}= 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]

Option A:
$9 \times 10^{19}$
Option B:
$9 \times 10^{20}$
Option C:
$14.94 \times 10^{20}$
Option D:
$14.94 \times 10^{19}$
Correct

Answer: D

Step-by-step solution

$h v=(4 \times 4.002-15.348) \times 1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2} v=14.94 \times 10^{19} \mathrm{kHz}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction