A) \[227.62\,\,amu\]
B) \[112.11\,\,amu\]
C) \[90.3\,\,amu\]
D) \[23.8\,\,amu\]
Correct Answer: A
Solution :
[a] By conservation of momentum, we have \[0={{\vec{P}}_{\alpha }}+{{\vec{P}}_{d}}\Rightarrow {{\vec{P}}_{\alpha }}=-{{\vec{P}}_{d}}\] or \[{{P}_{\alpha }}={{P}_{d}}=P\] The kinetic energy released in the process \[K={{K}_{\alpha }}+{{K}_{p}}=\frac{{{P}^{2}}}{2{{m}_{\alpha }}}+\frac{{{P}^{2}}}{2{{m}_{d}}}\] \[=\frac{{{P}^{2}}}{2{{m}_{\alpha }}}\left( 1+\frac{{{m}_{\alpha }}}{{{m}_{d}}} \right)=\frac{{{(h/\lambda )}^{2}}}{2{{m}_{\alpha }}}\left( 1+\frac{{{m}_{\alpha }}}{{{m}_{d}}} \right)\] After substituting the given values, we get \[K=6.25\,\,MeV\] If \[{{m}_{P}}\] is the mass of the parent nucleus, then \[K+({{m}_{\alpha }}+{{m}_{d}}){{c}^{2}}={{m}_{p}}{{c}^{2}}\] 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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