When ^238U nucleus originally at rest, decays by emitting an alpha particle having a speed u , the recoil speed of the residual nucleus is: - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

From the law of conservation of linear momentum: P_{ini t ia l} = P_{f ina l} Since the nucleus was at rest: 0 = m_{α} v_{α} + m_{res i d u a l} v_{reco i l} .Mass of alpha particle m_{α} = 4 .Mass of residual nucleus m_{res i d u a l} = 238 - 4 = 234 . 0 = 4 u + 234 v v = - \frac{4 u}{234} Magnitude of recoil speed is \frac{4 u}{234} .

Try it yourself before checking the explanation above.

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

\frac{4 u}{238}

- \frac{4 u}{234}

\frac{4 u}{234}

- \frac{4 u}{238}

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