If _{92}{U^{238}} undergoes successively 8 al | vedclass

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Question

$_{(Z=92)}{{U}^{(A=238)}}{{\xrightarrow{(8\alpha ,\,6\beta )}}_{{{Z}'}}}{{X}^{{{A}'}}}$

so $A' = A - 4{n_\alpha } = 238 -(4 	imes 8) =

Vedclass · Accepted Answer

$_{82}P{b^{206}}$

Vedclass · Answer

$_{(Z=92)}{{U}^{(A=238)}}{{\xrightarrow{(8\alpha ,\,6\beta )}}_{{{Z}'}}}{{X}^{{{A}'}}}$

so $A' = A - 4{n_\alpha } = 238 -(4 	imes 8) = 206$

and $Z' = {n_\beta } - 2{n_\alpha } + z = 6 -(2 	imes 8) + 92 = 82$

If $_{92}{U^{238}}$ undergoes successively $8 \alpha$- decays and $6 \beta$- decays, then resulting nucleus is

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