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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