A) \[-\left( \frac{4v}{234} \right)\]
B) \[\left( \frac{v}{234} \right)\]
C) \[\left( \frac{v}{238} \right)\]
D) \[\frac{4v}{238}\]
Correct Answer: A
Solution :
\[_{92}{{U}^{238}}{{\xrightarrow[{}]{{}}}_{90}}T{{h}^{234}}{{+}_{2}}H{{e}^{4}}\] The weight of remaining nucleus. From law of conservation \[{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}=0\] \[4v+234v'=0\] or \[v'=-\frac{4v}{234}\] where v' is rebound velocity of remaining nucleus.
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