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

### Question

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 $$931.5 \mathrm{MeVc}^{-2}$$.

(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.

(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. 

(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 $$10^{17}$$.

(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.

(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. 

Ans:

$(26.98434-26.98153) \times 931.5$
OR
$$2.6175 \propto \mathrm{MeV}$$ 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
OR
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 \begin{aligned} & N_{10}=9.50 \times 10^{15} \times e^{-0.00122 \times 600} \text { seen } \\ & N_{10}=4.57 \times 10^{15} \end{aligned}
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}

### Question

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

(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. 

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

Outline

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

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

(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.

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

Ans:

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

b( i)

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

ALTERNATIVE 1
$$0.40 \mathrm{~kg}$$ 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 $$\alpha \frac{1.2 \times 10^{26} \times 17.51 \times 10^6 \times 1.6 \times 10^{-19}}{0.4+0.6} \rrbracket=3.4 \times 10^{14} \ll \mathrm{J} \mathrm{kg}^{-1} » \checkmark$$
ALTERNATIVE 2
$$\alpha 17.51 \times 10^6 \times 1.6 \times 10^{-19}=» 2.8 \times 10^{-12} \ll \mathrm{J} »$$
AND
\begin{aligned} & \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} \end{aligned}

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