The number of $α$ and $β$ - particles emitted, when the following nuclear transformation takes place are x and y respectively. Determine the value of x and y.
$_{92}^{238} X$ $⟶$ $_{82}^{206} Y$ .

x = 8, y = 6

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x = 6, y = 8

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x = 8, y = 8

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x = 6, y = 6

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Solution

The correct option is C $x = 8, y = 6$
$_{92}^{238} X \to_{82}^{206} Y$

$_{92}^{238} X \to_{82}^{206} Y + 8 α + 6 β$

So, $x = 8, y = 6$

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