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A nucleus \(_Z{X^A}\) emits an \(\alpha \) particle with velocity \(v\). The recoil speed of the daughter nucleus is
1 \(\frac{{A - 4}}{{4v}}\)
2 \(\frac{{4v}}{{A - 4}}\)
3 \(v\)
4 \(\frac{v}{4}\)
Explanation:
According to law of conservation of linear momentum \(0 = {m_D}{v_D} + {m_\alpha }{v_\alpha }\)(\(\because \) The parent nucleus is initially at rest) Where subscripts \(D\) and \(\alpha \) represent daughter nucleus and \(\alpha \)-particle respectively \(0 = \left( {A - 4} \right){v_D} + 4v\) ( \(\because \) Mass of a nucleus is approximately proportional to its mass number) \({v_D} = - \frac{{4v}}{{\left( {A - 4} \right)}}\) Negative sign shows the recoil of the daughter nucleus
