Consider the following radioactive decay process:
${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow} A_1 \stackrel{\beta^{-}}{\longrightarrow} A_2 \stackrel{\gamma}{\longrightarrow} A_3 \stackrel{\alpha}{\longrightarrow} A_4 \stackrel{\beta^{+}}{\longrightarrow} A_5 \stackrel{\gamma}{\longrightarrow} A_6$
The mass number and the atomic number of $A_6$ are given by:

  • A
    $210$ and $82$
  • B
    $210$ and $84$
  • C
    $210$ and $80$
  • D
    $211$ and $80$

Explore More

Similar Questions

$A$ radioactive decay chain starts from $_{93}Np^{237}$ and produces $_{90}Th^{229}$ by successive emissions. The emitted particles can be:

What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay: $_{90}X^{200} \to _{80}Y^{168}$?

$1 \text{ Curie}$ is equal to

The ${ }_{92}^{238} U$ atom disintegrates to ${ }_{84}^{214} Po$ with a half-life of $4.5 \times 10^9$ years by emitting $6$ $\alpha$-particles and $n$ electrons. Here,$n$ is:

Alpha rays emitted from a radioactive substance 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