First slide

Nuclear stability

Question

The radio active nucleus ${}_{90}^{232}A$ decays into a stable nucleus ${}_{82}^{208}B$ .Then number of $α and β$ particles emitted are respectively.

Easy

A

$α = 3, β = 3$
B

$α = 6, β = 4$
C

$α = 6, β = 0$
D

$α = 4, β = 6$

Solution

Suppose number of emitted $α and β$ particles are x and y respectively.
$∴ 232 - 4 yx = 208 \Rightarrow x = 6 . Also, 90 - 2 x + y = 82 \Rightarrow 90 - 2 \times 6 + y = 82 \Rightarrow y = 4 .$

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