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Physics-

General

Easy

Question

An atomic nucleus $90 to the power of T h to the power of 232 end exponent end exponent$ emits several $alpha$ and $beta$ radiations and finally reduces to $80 to the power of P b to the power of 208 end exponent end exponent$ . It must have emitted

4 $α$ and 2 $β$
6 $α$ and 4 $β$
8 $α$ and 24 $β$
4 $α$ and 16 $β$

The correct answer is: 6 $alpha$ and 4 $beta$

$n subscript alpha end subscript equals fraction numerator A minus A ´ over denominator 4 end fraction equals fraction numerator 232 minus 208 over denominator 4 end fraction equals 6$
$n subscript beta end subscript equals 2 n subscript alpha end subscript minus Z plus Z ´ equals 2 cross times 6 minus 90 plus 82 equals 4$

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