Question 102: Atoms & Nuclei

Question

In a radioactive decay chain, $${}_{90}^{232}Th$$ nucleus decays to $${}_{82}^{212}Pb$$ nucleus. Let $${N_\alpha }$$ and $${N_\beta }$$ be the number of $$\alpha $$ and $${\beta ^ - }$$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

Accepted Answer

Given,
$${}_{90}^{232}Th\buildrel {} \over
 \longrightarrow {}_{82}^{212}pb$$
where N
$$\alpha$$
= number of $$\alpha$$ particles, N
$$\beta$$
= number of $$\beta$$ particles
We know that $$\alpha$$-decay is given as
$${}_Z^AX\buildrel {} \over
 \longrightarrow {}_{Z - 2}^{A - 4}Y + {\alpha ^ - }$$ particle
and $$\beta$$-decay is given as
$${}_Z^AX\buildrel {} \over
 \longrightarrow {}_{Z + 1}^AX + {\beta ^ - }$$ particle
In above decay process change in mass number A is 20 and change in atomic number Z is 8.
Therefore, change in mass number A $$\Rightarrow$$ number of $$\alpha$$-particles emitted are
$${N_\alpha } = {{20} \over 4} = 5$$
Now, emission of 5 $$\alpha$$ particles implies atomic number reduces by 10 but change in atomic number is 8, therefore, number of $$\beta$$-particles emitted is 2.
Therefore, N
$$\alpha$$
= 5 and N
$$\beta$$
= 2
So, option (A) and option (C) are correct.

Answer

$${N_\alpha } = 5$$

Answer

$${N_\alpha } = 6$$

Answer

$${N_\beta } = 2$$

Answer

$${N_\beta } = 4$$

Atoms & Nuclei

In a radioactive decay chain, ${}_{90}^{232}Th nucleus decays to {}_{82}^{212}Pb nucleus. Let {N_\alpha } and {N_\beta } be the number of \alpha and {\beta ^ - }$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

Choose the correct answer: