**Solutions: Radioactivity Questions**

**Q12. **The isotope of uranium, , decays into a stable isotope of lead, by means of a series of α and β– decays.

(a) In this series of decays, α decay occurs 8 times and β– decay occurs n times. Calculate n.

92 - 16 = 76 and

The final proton number is 82 - there are therefore 82 - 76

= 6 beta decays.

**(1 mark)**

(b)

(i) Explain what is meant by the binding energy of a nucleus.

The binding energy of a nucleus is the energy required to split the nucleus up into its **individual **neutrons and protons (or nucleons)nucleons

OR the energy released to form the nuleus from its **individual** neutrons and protons/nucleons

**(2 marks) **

(ii) The graph belowshows the binding energy per nucleon for some stable nuclides.

Estimate the binding energy, in MeV, of the nucleus.

From the graph the binding energy per nucleon is 7.88 MeV

We have 206 nucleons, so the total binding energy is 206 x 7.88 = 1620 MeV

**(1 mark)**

(c) The half-life of is 4.5 × 10^{9} years, which is much larger than all the other half-lives of the decays in the series. A rock sample when formed originally contained 3.0 × 10^{22} atoms of and no atoms.

At any given time most of the atoms are either or with a negligible number of atoms in other forms in the decay series.

(i) Using the axes below, sketch graphs to show how the number of atoms and the number of atoms in the rock sample vary over a period of 1.0 × 10^{10} years from its formation.

Label your graphs U and Pb.

U, a graph starting at 3.0 × 10^{22} showing exponential fall passing through 0.75 × 10^{22} at 9 × 10^{9} years

Pb, inverted graph of the above so that the graphs cross at 1.5 × 10^{22} at 4.5 × 10^{9} years

**(2 marks)**

(ii) A certain time, t, after its formation the sample contained twice as many atoms as atoms.

Show that the number of atoms in the rock sample at time t was 2.0 × 10^{22}.

Let number of uranium atoms be U and number of lead atoms be Pb

At time t

U + Pb = 3.0 × 10^{22}

But U = 2Pb

So, U + 0.5U = 3.0 × 10^{22}

U = 3.0 × 10^{22}/1.5 = 2.0 × 10^{22}

**(1 mark) **

(iii) Calculate t in years.

N = N_{o} e^{-λt}

2 × 10^{22} = 3 × 10^{22} × e^{-λt}

ln (2/3) =
-λt

ln (3/2) = ln 1.5 = λt

t = ln 1.5 / λ

**Now** λ = ln 2 / t_{1/2}

λ = ln 2 / 4.5 × 10^{9}

λ = 1.54 × 10^{-10}

t = ln 1.5/ 1.54 × 10^{-10}

t = 2.6 × 10^{9} years

**(3 marks)**

**(Total 10 marks)**