JEE PHYSICS MCQ: NUCLEI

1 Medium

1 of 1

Analyzing +4 | -1

{U^{238}}

decays into

T{h^{234}}

by the emission of an

\alpha -

particle. There follows a chain of further radioactive decays, either by

\alpha -

decay or by

\beta -

decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the end product of the

{U^{238}}

radioactive decay chain

View answer and explanation

Correct option: A

P{b^{206}}

Option A (

P{b^{206}}

) is correct because the (4n+2) series starts from

{U^{238}}

and its stable end product is

P{b^{206}}

JEE PHYSICS MCQ: NUCLEI

More Questions from NUCLEI