(D) The nuclear reaction is given by: $_{92}U^{235} + _{0}n^{1} \to _{54}Xe^{139} + _{38}Sr^{94} + x(_{0}n^{1})$.Applying the law of conservation of m

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(D) The nuclear reaction is given by: $_{92}U^{235} + _{0}n^{1} \to _{54}Xe^{139} + _{38}Sr^{94} + x(_{0}n^{1})$.Applying the law of conservation of mass number: $235 + 1 = 139 + 94 + x$.$236 = 233 + x$.$x = 3$.Therefore,the product $X$ is $3$ neutrons.

Class 12 Chemistry Nuclear Chemistry Artificial transmutation

Medium

If $_{92}U^{235}$ nucleus absorbs a neutron and disintegrates into $_{54}Xe^{139}$,$_{38}Sr^{94}$ and $X$,then what will be the product $X$?

A
$\alpha$-particle
B
$\beta$-particle
C
$2$ neutrons
D
$3$ neutrons

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Class 12

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Nuclear Chemistry

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Artificial transmutation

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Medium

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If $_{92}U^{235}$ nucleus absorbs a neutron and disintegrates into $_{54}Xe^{139}$,$_{38}Sr^{94}$ and $X$,then what will be the product $X$?

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Bombardment of aluminum by $\alpha$-particle leads to its artificial disintegration in two ways,$(i)$ and $(ii)$ as shown. Products $X,$ $Y$ and $Z$ respectively are,

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