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Q.

${}_{90}^{232}{Th}$ an isotope of thorium decays in ten stages emitting six $α$ -particles and four $β$ -particles in all. The end product of the decay is

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a

${}_{82}^{208}{Pb}$

b

${}_{90}^{232}{Th}$

c

${}_{82}^{206}{Pb}$

d

${}_{82}^{209}{Pb}$

answer is D.

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Detailed Solution

New mass number $A' = A - 4 n_{α} = 232 - 4 x 6 = 208$

Atomic number $Z' = Z + n_{β} - 2 n_{α} = 90 + 4 - 2 x 6 = 82$

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