If $_{92} U^{238}$ changes to $_{85} A t^{210}$ by a series $α -$ and $β - d e c a y s$ , the number of $α -$ and $β - d e c a y s$ undergone is

Question

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Answer 1

An alpha particle is a doubly charged Helium ion.
$α = H e^{2 +}$
Mass no. of $α$ = 4
Atomic no. of $α$ = 2

$β$ particle
Mass no. of $β$ = 0
Change in atomic no. Atomic no. by $β$ = $\pm 1$

Change in mass possible only by $α$ particles
$Δ M = 238 - 210 = 28$
No. of alpha decay = $\frac{2 8}{4} = 7$

Total no. of protons left = $92 - 2 \times 7 = 78$
Total protons in $_{85} A t^{210}$ = 85
So $β^{-}$ decay = 85-78 = 7
$n \to β^{-} + p$

Answer 2

An alpha particle is a doubly charged Helium ion.

α = H e^{2 +}

Mass no. of

α

= 4
Atomic no. of

α

= 2

β

particle
Mass no. of

β

= 0
Change in atomic no. Atomic no. by

β

=

\pm 1

Change in mass possible only by

α

particles

Δ M = 238 - 210 = 28

No. of alpha decay =

\frac{2 8}{4} = 7

Total no. of protons left =

92 - 2 \times 7 = 78

Total protons in

_{85} A t^{210}

= 85
So

β^{-}

decay = 85-78 = 7

n \to β^{-} + p

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