If _{92}U^{236} nucleus emits one alpha-parti | vedclass

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(B) The emission of an $\alpha$-particle $(_{2}He^{4})$ from a nucleus results in a decrease of $2$ in the atomic number $(Z)$ and $4$ in the mass num

Vedclass · Accepted Answer

$142$ neutrons and $90$ protons

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(B) The emission of an $\alpha$-particle $(_{2}He^{4})$ from a nucleus results in a decrease of $2$ in the atomic number $(Z)$ and $4$ in the mass number $(A)$.The nuclear reaction is: $_{92}U^{236} 	o _{90}X^{232} + _{2}He^{4}$.The number of protons $(Z)$ in the remaining nucleus is $92 - 2 = 90$.The number of neutrons $(N)$ is calculated as $A - Z = 232 - 90 = 142$.Therefore,the remaining nucleus has $142$ neutrons and $90$ protons.

If $_{92}U^{236}$ nucleus emits one $\alpha$-particle,the remaining nucleus will have

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