A) 1.0
B) 0.5
C) 0.25
D) 0.123
Correct Answer: D
Solution :
The initial momentum of parent nucleus is zero. Hence, the momenta of the emitted a-particle and of the daughter nucleus will be equal (and opposite) to each other (momentum-conservation), i.e., \[{{p}_{\alpha }}-{{p}_{d}}\] Kinetic energy of\[\alpha -\]particle \[{{K}_{\alpha }}=\frac{p_{\alpha }^{2}}{2m\alpha }\] = 6.7 MeV (Given) Kinetic energy or recoil energy of daughter nucleus \[{{K}_{d}}=\frac{1}{2}\frac{p_{d}^{2}}{{{m}_{d}}}\] \[\therefore \] \[\frac{{{K}_{d}}}{{{K}_{\alpha }}}=\left( \frac{{{p}_{d}}}{{{p}_{\alpha }}} \right)\frac{{{m}_{\alpha }}}{{{m}_{d}}}\Rightarrow \frac{{{K}_{d}}}{{{K}_{\alpha }}}=\frac{{{m}_{\alpha }}}{{{m}_{d}}}\] \[(\because pd=p\alpha )\] \[\therefore \] \[{{K}_{d}}={{K}_{\alpha }}.\frac{{{m}_{\alpha }}}{{{m}_{d}}}\] Now, putting \[{{K}_{\alpha }}=6.7\,\text{MeV,}\] \[{{m}_{\alpha }}=6.645\times {{10}^{-27}}kg\] and \[{{m}_{d}}=218\text{amu = 218}\times 1.66\times {{10}^{-27}}kg\] \[\therefore \] \[{{K}_{d}}=\frac{6.7\times 6.645\times {{10}^{-27}}}{218\times 1.66\times {{10}^{-27}}}\text{MeV}\] \[=0.123\,\text{MeV}\] \[\therefore \] Recoil energy of daughter nucleus \[=0.123\,\text{MeV}\]
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