A stationary nucleus of mass number 220 emits | vedclass

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(B) Let $m_1 = 216$ be the mass number of the daughter nucleus and $m_2 = 4$ be the mass number of the $\alpha$-particle.Let $k_1$ and $k_2$ be their

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$5.4$

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(B) Let $m_1 = 216$ be the mass number of the daughter nucleus and $m_2 = 4$ be the mass number of the $\alpha$-particle.Let $k_1$ and $k_2$ be their respective kinetic energies.According to the law of conservation of energy,$k_1 + k_2 = 5.5 \, MeV \dots (i)$According to the law of conservation of linear momentum,the magnitudes of momenta must be equal: $p_1 = p_2$.Since $p = \sqrt{2mk}$,we have $\sqrt{2(216)k_1} = \sqrt{2(4)k_2}$.Squaring both sides,$216 k_1 = 4 k_2$,which gives $k_2 = 54 k_1$.Substituting this into equation $(i)$,$k_1 + 54 k_1 = 5.5 \, MeV$.$55 k_1 = 5.5 \, MeV$,so $k_1 = 0.1 \, MeV$.Therefore,the kinetic energy of the $\alpha$-particle is $k_2 = 54 	imes 0.1 = 5.4 \, MeV$.

$A$ stationary nucleus of mass number $220$ emits an $\alpha$-particle. If the energy released in the process is $5.5 \, MeV$,find the kinetic energy of the $\alpha$-particle in $MeV$.

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