# IBDP Physics E.2 Quantum physics 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) Photons of wavelength $$468 \mathrm{~nm}$$ are incident on a metallic surface. The maximum kinetic energy of the emitted electrons is $$1.8 \mathrm{eV}$$.

Calculate

(i) the work function of the surface, in $$\mathrm{eV}$$.

(ii) the longest wavelength of a photon that will eject an electron from this surface.

(b) (i) In an experiment, alpha particles of initial kinetic energy $$5.9 \mathrm{MeV}$$ are directed at stationary nuclei of lead $$\left({ }_{82}^{207} \mathrm{~Pb}\right)$$. Show that the distance of closest approach is about $$4 \times 10^{-14} \mathrm{~m}$$.

(ii) The radius of a nucleus of $${ }_{82}^{207} \mathrm{~Pb}$$ is $$7.1 \times 10^{-15} \mathrm{~m}$$. Suggest why there will be no deviations from Rutherford scattering in the experiment in (b)(i).


Ans:

a i )Use of $$E_{\max }=\frac{h c}{\lambda}-\phi \Rightarrow \phi=\frac{h c}{\lambda}-E_{\max }$$
$\phi=\alpha \frac{h c}{\lambda}-E_{\max }=\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\left(468 \times 10^{-9}\right)\left(1.6 \times 10^{-19}\right)}-1.8 »=0.85625 \approx 0.86 \ll \mathrm{eV} \rightsquigarrow$

ii Use of $$\frac{h c}{\lambda}=\phi \Rightarrow \lambda=\frac{h c}{\phi}$$
$\lambda=\kappa \frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\left(468 \times 10^{-9}\right)\left(1.6 \times 10^{-19}\right)}=» 1.45 \times 10^{-5} \approx \mathrm{m} »$

b i 2e AND $$82 \mathrm{e}$$ seen
OR
$$3.2 \times 10^{-19} \approx \mathrm{C}$$ AND $$1.312 \times 10^{-17} ” \mathrm{C}$$ s seen $$\checkmark$$
$d=\frac{8.99 \times 10^9 \times(2 e)(82 e)}{5.9 \times 10^6 \times e}=3.998 \times 10^{-14} \approx 4 \times 10^{-14} \mathrm{\kappa mm}$

ii ) The closest approach is «significantly» larger than the radius of the nucleus / far away from the nucleus/OWTTE.
«Therefore the strong nuclear force will not act on the alpha particle.

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