JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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]

A
$9 \times 1 0^{19}$
B
$9 \times 1 0^{20}$
C
$14.94 \times 1 0^{20}$
D
$14.94 \times 1 0^{19}$

Solution

Mass defect: $Δ m = 4 m_{α} - m_{p a r t i c l e} = 4 (4.002) - 15.348 = 16.008 - 15.348 = 0.660$ amu.

Minimum photon energy must equal the mass-defect energy: $h ν = Δ m \cdot c^{2}$ .

Compute: $h ν = 0.660 \times 1.66 \times 1 0^{- 27} \times (3 \times 1 0^{8})^{2} = 0.660 \times 1.66 \times 9 \times 1 0^{- 11} \approx 9.86 \times 1 0^{- 11}$ J.

$ν = \frac{9.86 \times 1 0 ^{- 11}}{6.6 \times 1 0 ^{- 34}} \approx 1.494 \times 1 0^{23}$ Hz $= 1.494 \times 1 0^{20}$ kHz $= 14.94 \times 1 0^{19}$ kHz.

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