In the radioactive disintegration series 232 T h 90-208 P b 82- involving - and - decay- the total number of - and - particles emitted are-A- 6 - and 6 -B- 6 - and 5 -C- 5 - and 6 -D- 6 - and 4 -

The correct option is D 6α and 4βNo. of α particle = 232 2084=6 This is because nbsp;β decay does not reduce the mass number, so the loss in mass must be d ...

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Byju's Answer

Standard XII

Chemistry

Properties of Alpha, Beta and Gamma Rays

In the radioa...

Question

In the radioactive disintegration series $^{232} T h_{90} \to^{208} P b_{82}$ , involving $α a n d β$ decay, the total number of $α a n d β$ particles emitted are:

6 α a n d 6 β

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6 α a n d 4 β

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6 α a n d 5 β

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5 α a n d 6 β

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Solution

The correct option is B $6 α a n d 4 β$
No. of $α$ particle = $\frac{232 - 208}{4} = 6$

This is because $β$ decay does not reduce the mass number, so the loss in mass must be due to $α$ decay only.

If 6 Helium nuclei were emitted, then there should be $6 \times 2 = 12$ decrease in the atomic number, that is, we should have got a daughter nucleus having an atomic number of $90 - 12 = 78$ . The atomic number of the daughter nucleus is 82. So the difference is because of $β$ decays

No. of $β$ particle= 82-78=4

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In the radioactive disintegration series 232Th90→208Pb82, involving α and β decay, the total number of α and β particles emitted are:

In the radioactive disintegration series $^{232} T h_{90} \to^{208} P b_{82}$ , involving $α a n d β$ decay, the total number of $α a n d β$ particles emitted are: