IB DP Physics E.3 Radioactive decay IB Style Question Bank HL Paper 2


Magnesium-27 nuclei \(\left({ }_{12}^{27} \mathrm{Mg}\right)\) decay by beta-minus \(\left(\beta^{-}\right)\)decay to form nuclei of aluminium-27 (Al).

(a) Show, using the data, that the energy released in the decay of one magnesium-27 nucleus is about \(2.62 \mathrm{MeV}\).
Mass of aluminium-27 atom \(=26.98153 u\)
Mass of magnesium-27 atom \(=26.98434 u\)
The unified atomic mass unit is \(931.5 \mathrm{MeVc}^{-2}\).[1]

(b) A Magnesium-27 nucleus can decay by one of two routes:
Route 1: \(70 \%\) of the beta particles are emitted with a maximum kinetic energy of \(1.76656 \mathrm{MeV}\), accompanied by a gamma photon of energy \(0.84376 \mathrm{MeV}\).
Route 2: \(30 \%\) of the beta particles have a maximum kinetic energy of \(1.59587 \mathrm{MeV}\) with a gamma photon of energy \(1.01445 \mathrm{MeV}\).
The final state of the aluminium-27 nucleus is the same for both routes.

(i) State the conclusion that can be drawn from the existence of these two routes.[1]

(ii) Calculate the difference between the magnitudes of the total energy transfers in parts (a) and (b).

(iii) Explain how the difference in part (b)(ii) arises. [1]

(c) Small amounts of magnesium in a material can be detected by firing neutrons at magnesium-26 nuclei. This process is known as irradiation.

Magnesium-27 is formed because of irradiation. The products of the beta-particle emission are observed as the magnesium-27 decays to aluminium-27.

(i) The smallest mass of magnesium that can be detected with this technique is \(1.1 \times 10^{-8} \mathrm{~kg}\).
Show that the smallest number of magnesium atoms that can be detected with this technique is about \(10^{17}\).[2]

(ii) A sample of glass is irradiated with neutrons so that all the magnesium atoms become magnesium-27. The sample contains \(9.50 \times 10^{15}\) magnesium atoms.
The decay constant of magnesium-27 is \(1.22 \times 10^{-3} \mathrm{~s}^{-1}\).
Determine the number of aluminium atoms that form in 10.0 minutes after the irradiation ends.[3]

(iii) Estimate, in W, the average rate at which energy is transferred by the decay of magnesium-27 during the 10.0 minutes after the irradiation ends. [2]



(26.98434-26.98153) \times 931.5
\(2.6175 \propto \mathrm{MeV}\) s seen \(\checkmark\)

b i evidence for nuclear energy levels \(\checkmark\)

b ii. Difference \(=2.6175-(1.76656+0.84376)=2.6175-2.61032=0.007195 \propto \mathrm{MeV} »\) OR
Difference \(=2.6175-(1.59587+1.01445)=2.61032=0.007195 \& \mathrm{MeV} n \checkmark\)

iii Another particle/ «anti» neutrino is emitted «that accounts for this mass / energy» \(\checkmark\)

c i So \(1.1 \times 10^{-8} \mathrm{~kg} \equiv \frac{1.1}{0.027} \times 10^{-8}\) «mols
Mass of atom \(=27 \times 1.66 \times 10^{-27} \alpha \mathrm{kg} n \checkmark\)
\(2.4-2.5 \times 10^{17}\) atoms \(\checkmark\)

ii $
& N_{10}=9.50 \times 10^{15} \times e^{-0.00122 \times 600} \text { seen } \\
& N_{10}=4.57 \times 10^{15}
So number of aluminium-27 nuclei \(=(9.50-4.57) \times 10^{15}=4.9(3) \times 10^{15} \checkmark\)

iii  \(\begin{aligned} & \text { Total energy released }=\text { ans (c)(ii) } \times 2.62 \times 10^6 \times 1.6 \times 10^{-19} \ll=2100 \mathrm{~J} n \\ & \ll \frac{2100}{600}=\$ 3.4-3.5 \ll W »\end{aligned}\)


(a) Identify with ticks [ \(\checkmark\) ] in the table, the forces that can act on electrons and the forces that can act on quarks. [2]

(b) The following data is available for atomic masses for the fusion reaction
{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}:

(i) Show that the energy released is about 18MeV. [2]

(ii) Estimate the specific energy of hydrogen by finding the energy produced when \(0.4 \mathrm{~kg}\) of \({ }_1^2 \mathrm{H}\) and \(0.6 \mathrm{~kg}\) of \({ }_1^3 \mathrm{H}\) undergo fusion. [2]

(c) It is hoped that nuclear fusion can be used for commercial production of energy.


(i) two difficulties of energy production by nuclear fusion.[2]

(ii) one advantage of energy production by nuclear fusion compared to nuclear fission. [1]

(d) Tritium \(\left({ }_1^3 \mathrm{H}\right)\) is unstable and decays into an isotope of helium \((\mathrm{He})\) by beta minus decay with a half-life of 12.3 years.

(i) State the nucleon number of the \(\mathrm{He}\) isotope that \({ }_1^3 \mathrm{H}\) decays into.[1]

(ii) The following diagram is an incomplete Feynman diagram describing the beta minus decay of \({ }_1^3 \mathrm{H}\) into \(\mathrm{He}\). Complete the diagram and label all the missing particles.[3]

(iii) Estimate the fraction of tritium remaining after one year. [2]



a ) Weak nuclear: 2 ticks \(\checkmark\)
Strong nuclear: quarks only \(\checkmark\)

b( i) 

\varangle \mu »=2.0141+3.0160-(4.0026+1.008665) \ll=0.0188 u »
In MeV: \(1876.13415+2809.404-(3728.4219+939.5714475)\)
=0.0188 \times 931.5 \text { OR }=17.512 \alpha \mathrm{MeV} » \checkmark

\(0.40 \mathrm{~kg}\) of deuterium is \(\ll \frac{400}{2} \times 6.02 \times 10^{23} \mathrm{w}=1.2 \times 10^{26}\) nuclei
\({ }_\alpha 0.60 \mathrm{~kg}\) of tritium is the same number \({ }_* \checkmark\)
So specific energy \(\alpha \frac{1.2 \times 10^{26} \times 17.51 \times 10^6 \times 1.6 \times 10^{-19}}{0.4+0.6} \rrbracket=3.4 \times 10^{14} \ll \mathrm{J} \mathrm{kg}^{-1} » \checkmark\)
\(\alpha 17.51 \times 10^6 \times 1.6 \times 10^{-19}=» 2.8 \times 10^{-12} \ll \mathrm{J} »\)
& \ll(2.0141+3.0160) \times 1.66 \times 10^{-27}=» 8.35 \times 10^{-27} \\
& \ll \frac{2.8 \times 10^{-12}}{8.35 \times 10^{-37}},=3.4 \times 10^{14} \ll \mathrm{Jkg}^{-1}

c i Requires high temp/pressure
Must overcome Coulomb/intermolecular repulsion
Difficult to contain / control «at high temp/pressurew
Difficult to produce excess energy/often energy input greater than output / OWTTE \(\checkmark\)
Difficult to capture energy from fusion reactions
Difficult to maintain/sustain a constant reaction rate

 ii ) Plentiful fuel supplies \(O R\) larger specific energy \(O R\) larger energy density \(O R\) little or no «major radioactive» waste products

d i ). 3

ii ) Proton shown
W- shown
Produces electron \(/ \mathrm{e}^{-} / \beta^{-}\)and antineutrino \(/ \bar{v}\) with proper arrow directions.

iii ) \(\begin{aligned} & \lambda=\ll \frac{\ln 2}{12.3} \rightsquigarrow 0.056 \kappa y^{-1} » \text { OR } 0.5^{\frac{1}{m}} \text { OR } e^{-1 \times \frac{\ln 2}{12.3}} \\ & 0.945 \text { OR } 94.5 \%\end{aligned}\)

Scroll to Top