Question

In uranium series, what will be the decay constant of the stable end product of the series, if the decay constant of the parent nuclide is $\lambda $ ?

• A. $\frac{\lambda }{238}$
• B. $\frac{\lambda }{206}$
• C. $\frac{\lambda }{208}$
• D. zero

Answer 1

Answer: (D)

Rate of decay $\frac{d N}{d t}=-\lambda N$
where, $N \rightarrow numberofundecayednuclei\lambda \rightarrow decayconstant$
According to question , nucleus is stable.
so no decay will be there or $\frac{d N}{d t}=0-\lambda N=0\Rightarrow \lambda =0$