hv=(4×4.002−15.348)×1.66×10−27×(3×108)2
v=14.94×1019kHz
JEE Main 2026 — Physics Modern Physics
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles is ____ kHz.
[mass of He nucleus = 4.002amu,1amu=1.66×10−27 kg, h=6.6×10−34 J.s and c=3×108 m/s ]
Held on 22 Jan 2026 · Verified 6 Jul 2026.
14.94×1019
9×1020
14.94×1020
9×1019
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