Question

The ^{238}U nucleus has a relatively complicated
radioactive decay sequence until it becomes a stable
^{206}Pb 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 ^{238}U nucleus to decay into a stable
^{206}Pb nucleus?

1)How many protons and neutrons?

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

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 ^{206}Pb 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
^{238}U to ^{206}Pb.

**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 ^{238}U nucleus has a relatively complicated
radioactive decay sequence until it becomes a stable
^{206}Pb 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 ^{238}U nucleus to decay into a stable
^{206}Pb 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**

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