Question

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

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

Answer 1

The correct option is A $146$

Balance reaction is,
${}_{92}^{238}\text{M}{⟶}_{90}^{234}\text{N}{+}_2^4\text{He}$ and
${}_{90}^{234}\text{N}{⟶}_{88}^{234}\text{L}+2{\beta }^{+}\left(or2{}_{+1}^0\text{e}\right)$
so atomic mass of L is $234$ and no of protons in L is $88$
$\therefore Numberofneutron=234-88=146$