Question

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

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

Answer 1

Answer: (B)

$144$

The nuclear reactions are as given below.
${}_{92}^{238}\text{M}{\to }_{88}^{230}\text{N}+2{}_2^4\text{He};$
${}_{88}^{230}\text{N}{\to }_{86}^{230}\text{L}+2{\beta }^{+}$
The number of neutrons in the element L is $230-86=144$