JEE Main Physics question
A ²³⁸U nucleus decays by emitting an alpha particle of speed v ms⁻¹. The recoil speed of the residual nucleus is (in ms⁻¹)
Answer: (A)
Solution
By conservation of momentum (nucleus initially at rest, alpha particle mass 4, residual nucleus mass 234): 4v + 234V = 0 ⟹ V = −4v/234.