Q51 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
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 = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
[mass of He nucleus = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
- Correct answerA$14.94 \times 10^{19}$
- B$9 \times 10^{20}$
- C$14.94 \times 10^{20}$
- D$9 \times 10^{19}$
Solution
$h v =(4 \times 4.002-15.348) \times 1.66 \times 10^{-27} \times\left(3 \times 10^8\right)^2$
$v =14.94 \times 10^{19} \mathrm{kHz}$
$v =14.94 \times 10^{19} \mathrm{kHz}$
Hence, the correct answer is option: 'A'.