The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles is ______ kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66 × 10⁻²⁷ kg, h = 6.6 × 10⁻³⁴ J.s and c = 3 × 10⁸ m/s ] (a) 14.94 × 10²⁰ (b) 9 × 10¹⁹ (c) 9 × 10²⁰ (d) 14.94 × 10¹⁹

Answer: (d). Mass defect to supply: Δ m = 4(4.002) - 15.348 = 0.660 amu E = Δ m c² = 0.660 × 1.66 × 10⁻²⁷ × 9 × 10¹⁶ = 9.86 × 10⁻¹¹ J ν = E/h = 9.86 × 10⁻¹¹6.6 × 10⁻³⁴ = 1.494 × 10²³ Hz = 1.494 × 10²⁰ kHz = 14.94 × 10¹⁹ kHz So the answer is (d).

The minimum frequency of photon required to break a particle of mass… | JEE Main 2026 (22 Jan, Shift 1)

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Nuclear Reactions & Q-value

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 = $4.002\text{ amu}, 1\text{ amu} = 1.66 \times 10^{-27}\text{ kg}, \text{h} = 6.6 \times 10^{-34}\text{ J.s}$ and $\text{c} = 3 \times 10^{8}\text{ m/s}$ ]

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Answer: (d)

Mass defect to supply: $\Delta m = 4(4.002) - 15.348 = 0.660\ \text{amu}$
$E = \Delta m\,c^2 = 0.660 \times 1.66 \times 10^{-27} \times 9 \times 10^{16} = 9.86 \times 10^{-11}\ \text{J}$
$\nu = \dfrac{E}{h} = \dfrac{9.86 \times 10^{-11}}{6.6 \times 10^{-34}} = 1.494 \times 10^{23}\ \text{Hz}$
$= 1.494 \times 10^{20}\ \text{kHz} = 14.94 \times 10^{19}\ \text{kHz}$
So the answer is (d).

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