$_{90} T h^{228} ⟶_{84} B i^{212}$

Which among the given options is correct?

4 α, 1 β

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4 α, 2 β

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5 α, 1 β

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5 α, 2 β

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Solution

The correct option is B $4 α, 2 β$
$K e y p o i n t :$
In $α D e c a y$ mass number that is reduced by four and an atomic number that is reduced by two.
In $β d e c a y$ mass number remains the same and an atomic number is increased by one.

Here,

The number of $α - p a r t i c l e s$ is given by the formula:

$= \frac{mass number of emitter - mass number of end product}{4}$

$\frac{228 - 212}{4} = \frac{16}{4} = 4$

And,

$Number of β - p a r t i c l e s = 2 \times α - p a r t i c l e - (difference in atomic number)$

$= 2 \times 4 - (90 - 84) = 8 - 6 = 2$

Hence, $α$ and $β -$ particles, emitted are $4$ and $2$ respectively.

Option B is the correct answer

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