JEE Main202622 January 2026Morning ShiftPhysicsNuclear PhysicsActual
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 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 ]
Options
- A14.94 10¹⁹
- B9 10²⁰
- C14.94 10²⁰
- D9 10¹⁹
Correct answer
A. 14.94 10¹⁹
Step-by-step solution
h v =(4 4.002-15.348) 1.66 10⁻²⁷ (3 10^8 )^2 v =14.94 10¹⁹ kHz