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13.Nuclei
hard
In a radioactive decay chain reaction, ${ }_{90}^{230} Th$ nucleus decays into ${ }_{84}^{214} Po$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$particles emitted in this process is. . . . .
A
$4$
B
$2$
C
$3$
D
$8$
(IIT-2022)
Solution
$Th _{90}^{230} \rightarrow Po _{24}^{214}+ n _2^4+ m \beta_{-1}^0$
$230=214+4 n$
$n =\frac{16}{4}=4$
$90=84+ n \times 2- m \times 1$
$90=84+4 \times 2- m \times 1$
$m =92-90=2$
Hence$\frac{ n }{ m }=\frac{4}{2}=2$ Ans.
Hence $\frac{ n }{ m }=\frac{4}{2}=2$ Ans.
Standard 12
Physics
