Question

A nucleus of mass 218 amu in Free State decays to emit an a-particle. Kinetic energy of the a-particle emitted is 6.7 MeV. The recoil energy ( in MeV) of the daughter nucleus is:

• A. 1
• B. 0.5
• C. 0.25
• D. 0.123

Answer 1

The Correct Option is D

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} $