| Consider decay process. |
| \[_{92}^{232}U\xrightarrow{{}}_{92}^{228}Th+_{2}^{4}He\] |
| masses involved are |
| \[m\,({}^{232}U)=232.1095\,u\] |
| \[m\,({}^{228}Th)=228.0998\,u\] |
| \[m\,({}^{4}He)=4.0039\,u\] |
| Kinetic energy of emitted \[\alpha \]-particle is |
A) always\[5.30MeV\].
B) always less than \[5.30MeV\]
C) always greater than \[5.30MeV\]
D) either \[5.30MeV\] or less
Correct Answer: D
Solution :
| Using, \[dW=qV=\frac{{{z}^{2}}\rho }{3\in }4\pi {{z}^{2}}dz.\rho \] |
| we get |
| \[Q=5.40MeV\] |
| This energy is distributed between \[\alpha -\]particle and daughter nucleus. |
| Energy is shared in inverse proportion a masses. So, ratio of kinetic energies is \[\frac{{{E}_{k}}(Th)}{{{E}_{k}}(\alpha )}=\frac{4}{228}\] |
| So, \[{{E}_{k}}(Th)=0.10\,MeV\] |
| and \[{{E}_{k}}(\alpha )=5.30MeV\] |
| But if \[{}^{228}Th\]is in an excited state, then |
| \[{{E}_{k}}(\alpha )\] is less. |
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