| Consider the following reaction. |
| \[_{92}^{235}U+_{0}^{1}n\xrightarrow{{}}_{32}^{83}Gc+_{60}^{153}Nd\] |
| Now, the ratio of kinetic energies of \[Ge\] and \[Nd\] is |
A) \[83:153\]
B) \[1:1\]
C) \[11:6\]
D) \[60:32\]
Correct Answer: C
Solution :
| Both kinetic energy and linear momentum are conserved. |
| So, \[K{{E}_{Ge}}+K{{E}_{Nd}}=0\] and \[{{m}_{Ge}}{{v}_{Ge}}+{{m}_{Nd}}{{v}_{Nd}}=0\] |
| Combining both, we get |
| \[{{K}_{Ge}}=\left( \frac{{{m}_{Nd}}}{{{m}_{Nd}}+{{m}_{Ge}}} \right)\times Q\] |
| \[{{K}_{Nd}}=\left( \frac{{{m}_{Ge}}}{{{m}_{Nd}}+{{m}_{Ge}}} \right)\times Q\] |
| So, the ratio is \[\frac{{{K}_{Ge}}}{{{K}_{Nd}}}=\frac{{{m}_{Nd}}}{{{m}_{Ge}}}\]\[=\frac{153}{83}=\frac{11}{6}\] |
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