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]
▶️Answer/Explanation
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]
▶️Answer/Explanation
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.