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In the case of thorium (A = 232 and Z = ...

In the case of thorium `(A = 232 and Z = 90)`, we obtain an isotope of lead `(A = 208 and Z = 82)` after some radiactive disintegrations. The number of `alpha` and `beta` particle emitted are respectively.

A

6, 3

B

6, 4

C

5, 5

D

4, 6

Text Solution

Verified by Experts

The correct Answer is:
B
(b) Decrease in mass number
`=232 - 208 = 24`
Number of particles emitted
`=(24)/(6) = 6`
Due to emission of `6` particles, decrease in charge number `= 12`. But decrease in charge number is Clearly `4` particles are emitted.
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