In a radioactive decay chain reaction, 90230 Th nucleus decays into 84214 Po nucleus. The ratio of the number of α to number of β-particles emitted in this process is

Question

In a radioactive decay chain reaction, 90230 Th nucleus decays into 84214 Po nucleus. The ratio of the number of α to number of β-particles emitted in this process is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Th 90230 arrow Po 84214+ n 24+ m β-10 230=214+4 n n =(16/4)=4 90=84+ n × 2- m × 1 90=84+4 × 2- m × 1 m =92-90=2 Hence ( n / m )=(4/2)=2 Ans.

Q. In a radioactive decay chain reaction, $_{90}^{230} T h$ nucleus decays into $_{84}^{214} P o$ nucleus. The ratio of the number of $α$ to number of $β^{-}$ particles emitted in this process is _____

$T h_{90}^{230} \to P o_{84}^{214} + n_{2}^{4} + m β_{- 1}^{0}$
$230 = 214 + 4 n$
$n = \frac{16}{4} = 4$
$90 = 84 + n \times 2 - m \times 1$
$90 = 84 + 4 \times 2 - m \times 1$
$m = 92 - 90 = 2$
Hence $\frac{n}{m} = \frac{4}{2} = 2$ Ans.

Q. In a radioactive decay chain reaction, 90230​Th nucleus decays into 84214​Po nucleus. The ratio of the number of α to number of β−particles emitted in this process is _____

Th90230​→Po84214​+n24​+mβ−10​ 230=214+4n n=416​=4 90=84+n×2−m×1 90=84+4×2−m×1 m=92−90=2 Hence mn​=24​=2 Ans.

Q. In a radioactive decay chain reaction, $_{90}^{230} T h$ nucleus decays into $_{84}^{214} P o$ nucleus. The ratio of the number of $α$ to number of $β^{-}$ particles emitted in this process is _____

$T h_{90}^{230} \to P o_{84}^{214} + n_{2}^{4} + m β_{- 1}^{0}$
$230 = 214 + 4 n$
$n = \frac{16}{4} = 4$
$90 = 84 + n \times 2 - m \times 1$
$90 = 84 + 4 \times 2 - m \times 1$
$m = 92 - 90 = 2$
Hence $\frac{n}{m} = \frac{4}{2} = 2$ Ans.