Question

${}_{98}^{238}X\stackrel{-\alpha }{\to }Y\stackrel{-\beta }{\to }Z\stackrel{-\beta }{\to }L\stackrel{-n\alpha }{\to }{}_{90}^{218}M$

In the sequence of the given nuclear reaction, what is the value of $n$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is B 4

• Alpha decay or $\alpha -decay$ is a type of radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by two.

• And beta decay $(\beta -decay)$ is a type of radioactive decay in which a beta particle (fast energetic electron or positron) is emitted from an atomic nucleus, transforming the original nuclide to an isobar.

According to question

${}_{98}^{238}X\stackrel{-\alpha }{\to }{}_{96}^{234}Y{+}_2^{4}He\stackrel{-\beta }{\to }{}_{97}^{234}Z{+}_{-1}^0\beta \stackrel{-\beta }{\to }{}_{98}^{234}L{+}_{-1}^0\beta \stackrel{-n\alpha }{\to }{}_{90}^{218}M+4{}_2^{4}He$

Hence , $n=4.$

therefore, option B is correct.