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

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Modern Physics

The minimum frequency of photon required to break a particle of mass 15.348 amu into

4 α

particles is ______ kHz.
[mass of He nucleus = 4.002 amu,

1

amu

= 1.66 \times 1 0^{- 27}

kg,

h = 6.6 \times 1 0^{- 34}

J.s and

c = 3 \times 1 0^{8}

m/s]

Answer: (D)

Splitting the particle into four

α

-particles requires supplying the mass defect

Δ m = 4 (4.002) - 15.348 = 0.66 amu

. The minimum photon energy equals this rest-energy gap,

h ν = Δ m c^{2} = 0.66 \times 1.66 \times 1 0^{- 27} \times (3 \times 1 0^{8})^{2} = 9.86 \times 1 0^{- 11} J

. Hence

ν = \frac{9.86 \times 1 0 ^{- 11}}{6.6 \times 1 0 ^{- 34}} = 1.494 \times 1 0^{23} Hz = 14.94 \times 1 0^{19} kHz .

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option