# IB DP Physics E.1 Structure of the atom IB Style Question Bank HL Paper 2

### 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]

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]

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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