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In the given nuclear reaction A, B, C, D...

In the given nuclear reaction `A, B, C, D, E` represents `._92 U^238 rarr^(alpha) ._BTh^A rarr^(beta) ._D Pa^C rarr^(E) ._92 U^234`.

A

`A=234,B=90,C=234,D=91andE=beta`

B

`A=234,B=90,C=238,D=94andE=alpha`

C

`A=238,B=93,C=234,D=91andE=beta`

D

`A=234,B=90,C=234,D=93andE=alpha`

Text Solution

Verified by Experts

The correct Answer is:
A
`""_(92)U^(238)overset(alpha)to_(90)Th^(234)overset(beta)to_(91)Pa^(234)overset(""_(-1)beta^(0))to_(92)U^(234)`
So, `A= 234 B= 90, C= 234 D = 91 and E=beta`
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