Question

A radioactive atom ${}_Y^{X}M$ emits two $\alpha$-particles and one $\beta$-particle respectively. The number of neutrons in the product nucleus will be :

• A. $X-Y-4$
• B. $X-5-Y$
• C. $X-3-Y$
• D. $X-Y-6$

Answer 1

Answer: (B)

$X-5-Y$

${}_Y^{X}M{⟶}_Q^{P}A+2_2^4\text{He}{+}_{-1}^0\text{e}$
Balancing mass number, we get
$X=P+8$
$\therefore P=X-8$

Balancing atomic number, we get
$Y=Q+4-1$
$\therefore Q=Y-3$
Number of neutrons in the product nucleus$=P-Q$
$=X-8-Y+3$
$=X-5-Y$