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If .92 U^238 undergoes successively 8 al...

If `._92 U^238` undergoes successively `8 alpha-`decays and `6 beta-`decays, then resulting nucleus is.

A

`._82 U^206`

B

`._86 Pb^206`

C

`._82 U^210`

D

`._82 U^214`

Text Solution

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The correct Answer is:
B
(b) `._((Z = 92)) U^((A = 238))overset (((8 alpha, 6 beta)))rarr ._Z' X^A'`
so `A' = A - 4 n_alpha = 238 - 4 xx 8 = 206`
and `Z' = n beta - 2 n_alpha + z = 6 - 2 xx 8 + 92 = 82`.
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