The isotope 86222Rn disintegrates by a decay series of - and - emissions to its conclusion at 82206Pb- What are the numbers of - and - particles lost respectively-

The correct option is A 4 and 4 The isotope 22286Rn disintegrates by a decay series of #945; and #946; emissions #10;to its conclusion at 20 ...

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Standard XII

Chemistry

Properties of Alpha, Beta and Gamma Rays

The isotope ...

Question

The isotope $_{86}^{222} R n$ disintegrates by a decay series of $α$ and $β$ emissions to its conclusion at $_{82}^{206} P b$ . What are the numbers of $α$ and $β$ particles lost respectively?

4

and

4

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4

and

8

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6

and

2

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8

and

4

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Solution

The correct option is A $4$ and $4$
The isotope 22286Rn disintegrates by a decay series of α and β emissions to its conclusion at 20682Pb. 4 and 4 are the numbers of α and β particles lost respectively.

A series of regular disintegration starting from an unstable nucleus and ending at a stable nucleus, is known as radioactive disintegration series. Or All the disintegration steps involved in the formation of a non-radioactive element from a radioactive element constitute the disintegration series.

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The isotope 22286Rn disintegrates by a decay series of α and β emissions to its conclusion at 20682Pb. What are the numbers of α and β particles lost respectively?

The isotope $_{86}^{222} R n$ disintegrates by a decay series of $α$ and $β$ emissions to its conclusion at $_{82}^{206} P b$ . What are the numbers of $α$ and $β$ particles lost respectively?