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.

6, 3

6, 4

5, 5

4, 6

Text Solution

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The correct Answer is:

(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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