In the uranium series,the decay constant of the parent nuclide is $\lambda$. What is the decay constant of the stable end product of this series?

  • A
    $\lambda / 238$
  • B
    $\lambda / 206$
  • C
    $\lambda / 208$
  • D
    Zero

Explore More

Similar Questions

In the nuclear reaction ${}_{92}^{235}U$ decaying to ${}_{91}^{231}Pa$,what are the particles emitted?

Difficult
View Solution

$A$ radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3$ $\alpha$-particles and $2$ positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be:

If a nucleus emits an $e^-$ particle,then its neutron-to-proton ratio $[n/p]$ will .....

In the uranium radioactive series,the initial nucleus ${}^{238}_{92}U$ decays to the final nucleus ${}^{206}_{82}Pb$. In this process,the number of $\alpha$-particles and $\beta$-particles emitted are:

Consider a $\beta$ decay reaction:
${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$
The atomic masses of ${}_1^3H$ and ${}_2^3He$ are $3.016050 \, u$ and $3.016030 \, u$,respectively. Find the maximum possible energy of the electron in $MeV$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo