SATHEE: Atoms And Nuclei Question 405

Atoms And Nuclei Question 405

Question: A nucleus at rest undergoes a decay emitting an $α -$ particle of de-Broglie wavelength $λ = 5.76 \times 10^{- 15} m .$ If the mass of the daughter nucleus is $223.610$ amu and that of the $α -$ particle is $4.002, a m u,$ determine the mass of the parent nucleus inamu. $(1, a m u = 931.470 M e V / c^{2}] .$

Options:

A) $227.62 a m u$

B) $112.11 a m u$

C) $90.3 a m u$

D) $23.8 a m u$

Show Answer

Answer:

Correct Answer: A

Solution:

By conservation of momentum, we have

$0 = {\vec{P}}_{α} + {\vec{P}}_{d} \Rightarrow {\vec{P}}_{α} = - {\vec{P}}_{d}$

or $P_{α} = P_{d} = P$ The kinetic energy released in the process $K = K_{α} + K_{p} = \frac{P^{2}}{2 m_{α}} + \frac{P^{2}}{2 m_{d}}$ $= \frac{P^{2}}{2 m_{α}} (1 + \frac{m_{α}}{m_{d}}) = \frac{{(h / λ)}^{2}}{2 m_{α}} (1 + \frac{m_{α}}{m_{d}})$

After substituting the given values, we get $K = 6.25 M e V$

If $m_{P}$ is the mass of the parent nucleus, then $K + (m_{α} + m_{d}) c^{2} = m_{p} c^{2}$ or $6.25 + (223.61 + 4.002) c^{2} = m_{p} c^{2}$

After simplifying, we get $m_{p} = 227.62$ amu.

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