Nuclei Ques 93
93. What is the respective number of $\alpha$ and $\beta$-particles emitted in the following radioactive decay ${ }^{200} X _{90} \to{ }^{168} Y _{80}$ ?
[1995]
(a) 6 and 8
(b) 6 and 6
(c) 8 and 8
(d) 8 and 6
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Answer:
Correct Answer: 93.(d)
Solution:
- (d) ${ }^{200} X _{90} \longrightarrow{ }^{168} Y _{80}$. We know that
${ }^{200} X _{90} \longrightarrow n_2 He^{4}+m _{-1} \beta^{0}+{ }^{168} Y _{80}$.
Therefore, in this process, $200=4 n+168$ or $n=\frac{200-168}{4}=8$.
Also, $90=2 n-m+80$
or, $m=2 n+80-90=(2 \times 8+80-90)=6$.
Thus, respective number of $\alpha$ and $\beta$-particles will be 8 and 6 .
BETA
