Nuclei - Result Question 46
47. Complete the equation for the following fission process : $ _{92} U^{235}+ _0 n^{1} \to _{38} Sr^{90}+\ldots$ [1998]
(a) $ _{54} X^{143}+3 _0 n^{1}$
(c) $ _57^{54} X^{142}+3 _0 n^{1}$
(b) $ _{54} X^{145}+3 _0 n^{1}$
(d) $ _{54} X^{142}+ _0 n^{1}$
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Answer:
Correct Answer: 47. (a)
Solution:
- (a) $ _{92} U^{235}+ _0 n^{1} \to _{38} Sr^{90}+ _{54} Xe^{143}$
$+3 n^{1}+$ energy
