A stationary Pb200 nucleus emits an -particle with certain kinetic energy K- If recoil velocity of daughter nucleus is 3-39 - 105 m-s and f is the fraction of the total energy liberated in this decay which is accounted for the recoil energy of the daughter nuclei- then

P_d=P_α =√(2m_αk_α) ⇒v_d=√(2m_αk_α)m_d Also, k_dK_α= m_αm_d K_dK_α+K_d=m_αm_d+m_α K_dK_T=( m_αm_α+m_d) nbsp; ...

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Second Law of Motion

A stationary ...

Question

A stationary $P b^{200}$ nucleus emits an $α -$ particle with certain kinetic energy $K_{α}$ . If recoil velocity of daughter nucleus is $3.39 \times 105 m / s$ and $f$ is the fraction of the total energy liberated in this decay which is accounted for the recoil energy of the daughter nuclei, then

K_{α} \approx 8.77 M e V

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f \approx 0.02

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f \approx 0.04

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K_{α} \approx 5.77 M e V

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