Question

The ²³⁸U nucleus has a relatively complicated radioactive decay sequence until it becomes a stable ²⁰⁶Pb nucleus.
The questions below should help you understand why this sequence is not a simple series of alpha decays, but involves both alpha and beta-minus decays.
How many protons and how many neutrons must be removed from the unstable ²³⁸U nucleus to decay into a stable ²⁰⁶Pb nucleus?

1)How many protons and neutrons?

2) Find the minimum, actual, and extra number of alpha decays for the radioactive decay of ²³⁸U to ²⁰⁶Pb.

a) Number of minimum alpha decays

b) Number of actual alpha decays

c) Number of "extra" alpha decays (= Actual number - Minimum number)

3)

These extra alpha decays occur because otherwise the ²⁰⁶Pb nucleus has too many neutrons and is not stable.
An extra alpha decay can occur after (choose one of the above)

  one beta minus (or electron) decay.
  one beta plus (or positron) decay.
  two beta minus (or electron) decays.
  two beta plus (or positron) decays.

4)

How many total beta-minus decays must occur to allow the extra alpha decays?

5)

So, let us summarize what happens for the radioactive decay of ²³⁸U to ²⁰⁶Pb.
Summary of PROTONS:

a) Number of protons removed due to alpha decays

b) Number of protons ADDED due to beta-minus decays

c) NET TOTAL number of protons removed

6)

Summary of NEUTRONS:

a)

Number of neutrons removed due to alpha decays

b) Number of neutrons removed due to beta-minus decays

c) NET TOTAL number of neutrons removed

Answer 1

The ²³⁸U nucleus has a relatively complicated radioactive decay sequence until it becomes a stable ²⁰⁶Pb nucleus.
The questions below should help you understand why this sequence is not a simple series of alpha decays, but involves both alpha and beta-minus decays.
How many protons and how many neutrons must be removed from the unstable ²³⁸U nucleus to decay into a stable ²⁰⁶Pb nucleus?

1)How many protons and neutrons?

Answer = 10 protons and 22 neutrons

steps

alpha decay, it is reduced by 4, with N and Z each being reduced by 2.

A=238 to A=206, A is reduced by 238-206=32, and that requires 8 α decays.

238U to 206 Pb decay

number alpha ,A changes by 32, there must be 8 alpha decays .

. These 8 alpha decays would decrease Z by 16units, from 92 to 76 int he chain.

Final Z must be 82, so there must also be 6 beta decays in the chain