_{92}U^{238} emits 8 alpha-particles and 6 be | vedclass

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Question

(C) The nuclear reaction is: $_{92}U^{238} \xrightarrow{-8\alpha, -6\beta} _{Z}X^{A}$.Change in mass number $(A)$: $238 - (8 	imes 4) = 238 - 32 = 20

Vedclass · Accepted Answer

$62/41$

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(C) The nuclear reaction is: $_{92}U^{238} \xrightarrow{-8\alpha, -6\beta} _{Z}X^{A}$.Change in mass number $(A)$: $238 - (8 	imes 4) = 238 - 32 = 206$.Change in atomic number $(Z)$: $92 - (8 	imes 2) + (6 	imes 1) = 92 - 16 + 6 = 82$.The product nucleus is $_{82}X^{206}$.Number of protons $(p)$ $= 82$.Number of neutrons $(n)$ $= A - Z = 206 - 82 = 124$.Neutron/proton ratio $= n/p = 124/82 = 62/41$.

$_{92}U^{238}$ emits $8$ $\alpha$-particles and $6$ $\beta$-particles. The neutron/proton ratio in the product nucleus is

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