The minimum frequency of photonINUCLEAR PHYSICS|JEE MAIN

JEE MAIN 2026

22-01-2026 S1

Question

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

Select the correct option:

$9 \times 10^{19}$

$9 \times 10^{20}$

$14.94 \times 10^{20}$

$14.94 \times 10^{19}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

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

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