Question

1. Mass number of a radioactive clement is
232 and its atomic number is 90. When
this element undergoes certain nuclear
reactions, it transforms into an isotope of
lead with a mass number 208 and an atomic
number 82. Determine the number of alpha
and beta decay that can occur.​

Answer 1

Answer:

$6 \alpha - decays \: and \: 4 { \beta }^{ - } \: - decays$

Explanation:

During alpha decay, the mass number decreases by 4 and the atomic number decreases by 2. Now, the element had final mass number 208 which means it decreased by,

232-208=2

So, the number of alpha decays should be 6 as 4×6=24. Also, the atomic number after 6 alpha decays would be,

90-12=78

But, the final atomic number is 82 and we know that during beta-minus decay the atomic number increases by 1 while the mass number remains unchanged. Hence, the number of beta-minus decays should be 4 as 4×1=1.

Hope it helps,

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