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Question

Physics Question on Nuclei

When U238U^{238} nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

A

4u238\frac{4u}{238}

B

4u234-\frac{4u}{234}

C

4u234\frac{4u}{234}

D

4u238-\frac{4u}{238}

Answer

4u238-\frac{4u}{238}

Explanation

Solution

According to principle of conservation of linear momentum the momentum of the system remains the same before and after the decay. Atomic mass of uranium = 238 and after emitting an alpha particle =2384=234= 238 - 4 = 234 ?238?0=4u+234v? 238 ? 0 = 4u + 234 v v=4u234\therefore v=-\frac{4u}{234}