In the nuclear decay sequence given below:
$_Z{X^A} \to {}_{Z + 1}{Y^A} \to {}_{Z - 1}{K^{A - 4}} \to {}_{Z - 1}{K^{A - 4}}$
the particles emitted in the sequence are:

  • A
    $\alpha, \beta, \gamma$
  • B
    $\beta, \alpha, \gamma$
  • C
    $\gamma, \alpha, \beta$
  • D
    $\beta, \gamma, \alpha$

Explore More

Similar Questions

The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} U \rightarrow{ }_{82}^{214} Pb$ is

In radioactive decay process,the negatively charged emitted $\beta$-particles are

Statement-$1$: $A$ nucleus having energy $E_{1}$ decays by $\beta^{-}$ emission to a daughter nucleus having energy $E_{2}$,but the $\beta^{-}$ rays are emitted with a continuous energy spectrum having an end-point energy $E_{1} - E_{2}$.
Statement-$2$: To conserve energy and momentum in $\beta$ decay,at least three particles must take part in the transformation.

$A$ common example of alpha decay is ${ }_{92}^{238} U \longrightarrow{ }_{90}^{234} Th +{ }_{2}^{4} He + Q$. (Given: ${ }_{92}^{238} U = 238.05060 \, u$,${ }_{90}^{234} Th = 234.04360 \, u$,${ }_{2}^{4} He = 4.00260 \, u$,and $1 \, u = 931.5 \, MeV/c^2$). The energy released $(Q)$ during the alpha decay of ${ }_{92}^{238} U$ is $...... \, MeV$.

$A$ radioactive nucleus ${}_{Z}^{A}X$ undergoes spontaneous decay in the sequence ${}_{Z}^{A}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$,where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are:

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