Question

If ${}_{92}{U}^{238}$ changes to ${}_{85}A{t}^{210}$ by a series $\alpha -$ and $\beta -decays$, the number of $\alpha -$ and $\beta -decays$ undergone is

• A. $7$ and $5$
• B. $7$ and $7$
• C. $5$ and $7$
• D. $7$ and $9$

Answer 1

Answer: (B)

$7$ and $7$

An alpha particle is a doubly charged Helium ion.
$\alpha =H{e}^{2+}$
Mass no. of $\alpha$ = 4
Atomic no. of $\alpha$ = 2
$\beta$ particle
Mass no. of $\beta$ = 0
Change in atomic no. Atomic no. by $\beta$ = $\pm 1$
Change in mass possible only by $\alpha$ particles
$\Delta M=238-210=28$
No. of alpha decay = $\frac{28}{4}=7$
Total no. of protons left = $92-2\times 7=78$
Total protons in ${}_{85}A{t}^{210}$ = 85
So ${\beta }^{-}$ decay = 85-78 = 7
$n\to {\beta }^{-}+p$