In the nuclear reaction _{92}U^{238} to _{82} | vedclass

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Question

(A) For the given nuclear reaction: $_{92}U^{238} 	o _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$$1$. Balancing the mass number: $238 = 206 + 4

Vedclass · Accepted Answer

$8, 6$

Vedclass · Answer

(A) For the given nuclear reaction: $_{92}U^{238} 	o _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$$1$. Balancing the mass number: $238 = 206 + 4x + 0y \implies 4x = 32 \implies x = 8$.$2$. Balancing the atomic number: $92 = 82 + 2x - y \implies 92 = 82 + 2(8) - y \implies 92 = 82 + 16 - y \implies 92 = 98 - y \implies y = 6$.Therefore,the values are $x = 8$ and $y = 6$.

In the nuclear reaction $_{92}U^{238} \to _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$,the values of $x$ and $y$ are respectively......

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