Question
When tritium \(\left({ }_1^3 \mathrm{H}\right)\) decays by beta-minus \(\left(\beta^{-}\right)\)decay, one of the products is a stable isotope of helium \((\mathrm{He})\).
(a) Outline what is meant by an isotope.[1]
(b) Identify, for the helium isotope produced in the tritium decay, its
(i) mass number.[1]
(ii) proton number.[1]
(c) Outline the quark change that occurs during this decay.[1]
(d) Compare the properties of the strong nuclear force and of the electromagnetic force that allow the helium nucleus to be stable.[3]
(e) A beta-minus particle and an alpha particle have the same initial kinetic energy.
Outline why the beta-minus particle can travel further in air than the alpha particle.[2]
▶️Answer/Explanation
Ans:
«An atom with OR
Same chemical properties \(A N D\) different physical properties \(\checkmark\)
\(\begin{aligned} & 3 r \\ & 2 r\end{aligned}\)
b i \(3 r\)
b ii \(2 \checkmark\)
c. \(\mathrm{d} \rightarrow \mathrm{u} \checkmark\)
d.Strong force is short range \& electromagnetic force is long range
Strong force is attractive between nucleons/neutrons \& protons
electromagnetic force is repulsive between protons \(\checkmark\)
Overall, the strong force dominates
e.Alphas have double charge «and so are better ionisers \(n \checkmark\)
alphas have more mass and therefore slower «for same energy»
so longer time/more likely to interact with the «atomic» electrons/atoms «and therefore better ionisers
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.