A nucleus with $Z = 92$ emits the following in a sequence : ${\rm{\alpha }},{\rm{\alpha }},{{\rm{\beta }}^ - },{{\rm{\beta }}^ - },{\rm{\alpha }},{\rm{\alpha }},{\rm{\alpha }},{\rm{\alpha }}:{{\rm{\beta }}^ - },{{\rm{\beta }}^ - },{\rm{\alpha }},{{\rm{\beta }}^ + },{{\rm{\beta }}^ + },{\rm{\alpha }}$. The $Z$ of the resulting nucleus is

JEE PHYSICS MCQ: NUCLEI

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A nucleus with

Z = 92

emits the following in a sequence :

\mathrm{\alpha },\mathrm{\alpha },{\mathrm{\beta }^ - },{\mathrm{\beta }^ - },\mathrm{\alpha },\mathrm{\alpha },\mathrm{\alpha },\mathrm{\alpha }:{\mathrm{\beta }^ - },{\mathrm{\beta }^ - },\mathrm{\alpha },{\mathrm{\beta }^ + },{\mathrm{\beta }^ + },\mathrm{\alpha }

. The

Z

of the resulting nucleus is

View answer and explanation

Correct option: B

Option B (78) is correct because the atomic number of the resulting nucleus is calculated by accounting for the changes due to alpha and beta emissions.

JEE PHYSICS MCQ: NUCLEI

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