Home / IBDP Physics E.2 Quantum physics IB Style Question Bank : HL Paper 2

# 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 $$931.5 \mathrm{MeVc}^{-2}$$.[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]

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 $$\mathrm{eV}$$.[2]

(ii) the longest wavelength of a photon that will eject an electron from this surface.[2]

(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 $$4 \times 10^{-14} \mathrm{~m}$$.[2]

(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).
[2]

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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