Question

Consider the following nuclear reactions:
${}_{92}^{238}M\to$ ${}_Y^{X}N+2_2^{4}He$; ${}_Y^{X}N⟶$ ${}_B^{A}L+2{\beta }^{+}$
The number of neutrons in the element $L$ is:

• A. $142$
• B. $144$
• C. $140$
• D. $146$

Answer 1

Answer: (B)

$144$

$\alpha$-decay

${}_Z^{A}X\underrightarrow{-\alpha }{}_{Z-2}^{A-4}Y$

$\beta$-decay

${}_Z^{A}X\underrightarrow{-{\beta }^{+}}{}_{Z-1}^{A}Y$

Now,

${}_{92}^{238}M\underrightarrow{-2\alpha }{}_{92-4}^{238-8}N$

${}_{88}^{230}N\underrightarrow{-{2\beta }^{+}}{}_{86}^{230}L$

No. of neutrons in L = 230-86 = 144.

$\therefore$144 neutrons are present in the nucleus of L.