Bohr modified the Rutherford model by introducing the condition mvr = n\(\frac{h}{{2\pi }}\). Outline the reason for this modification.

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

\[v = \sqrt {\frac{{k{e^2}}}{{{m_{\text{e}}}r}}} \]

where k is the Coulomb constant.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

\[r = \frac{{{h^2}}}{{4{\pi ^2}k{m_{\text{e}}}{e^2}}}\]

Calculate the electron’s orbital radius in (c)(ii).

Explain what may be deduced about the energy of the electron in the β^– decay.

Suggest why the β^– decay is followed by the emission of a gamma ray photon.

Calculate the wavelength of the gamma ray photon in (d)(ii).

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n\(\frac{h}{{2\pi }}\)»

[3 marks]

\(\frac{{{m_{\text{e}}}{v^2}}}{r} = \frac{{k{e^2}}}{{{r^2}}}\)

OR

KE = \(\frac{1}{2}\)PE hence \(\frac{1}{2}\)m_ev² = \(\frac{1}{2}\frac{{k{e^2}}}{r}\)

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

combining v = \(\sqrt {\frac{{k{e^2}}}{{{m_{\text{e}}}r}}} \) with m_evr = \(\frac{h}{{2\pi }}\) using correct substitution

«eg \({m_e}^2\frac{{k{e^2}}}{{{m_{\text{e}}}r}}{r^2} = \frac{{{h^2}}}{{4{\pi ^2}}}\)»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

« r = \(\frac{{{{(6.63 \times {{10}^{ - 34}})}^2}}}{{4{\pi ^2} \times 8.99 \times {{10}^9} \times 9.11 \times {{10}^{ - 31}} \times {{(1.6 \times {{10}^{ - 19}})}^2}}}\)»

r = 5.3 × 10^–11 «m»

[1 mark]

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

Photon energy

E = 0.48 × 10⁶ × 1.6 × 10^–19 = «7.68 × 10^–14 J»

λ = «\(\frac{{hc}}{E} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{7.68 \times {{10}^{ - 14}}}}\) =» 2.6 × 10^–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

Explain why photoelectrons are not emitted from the metal surface unless the frequency of incident light exceeds a minimum value.

identify the minimum value of the frequency \({f_0}\) for photoelectrons to be emitted.

determine the Planck constant.

calculate the work function, in eV, for the metal surface.

The student repeats the experiment with a different metal surface that has a smaller value

for the work function. On the graph in (e), draw a line to show how \({E_{\max }}\) varies with \(f\).

light consists of photons with energy \(E = hf\);

there is a minimum energy/work function required for electron to leave a particular metal;

\(hf\) must be larger than this value;

(value of intercept on \(f\)-axis =) \(9.3 \times 10^{14}\) Hz;

Allow answers in the range of 9.0 to 9.5, but with correct power of ten.

attempted use of gradient \(\left( {\frac{h}{e}} \right)\) to evaluate \(h\);

eg \(\frac{{2.4}}{{6 \times {{10}^{14}}}} = 4 \times {10^{ - 15}}{\text{ eVs}}\); (allow answers in the range of 3.9 to \(4.2 \times {10^{ - 15}}\))

\(6.4 \times {10^{ - 34}}{\text{ Js}}\); (allow ECF, eg answers in the range of 6.2 to \(6.7 \times {10^{ - 34}}\))

Do not allow a bald answer of \(6.6 \times {10^{ - 34}}\) as value is known.

or

use of y-intercept \( = \frac{{ - h{f_0}}}{e}\);

\(h = \frac{{3.77 \times 1.6 \times {{10}^{ - 19}}}}{{0.93 \times {{10}^{15}}}}\); (allow intercept between –3.6 and –3.8 eV)

\(6.4 \times {10^{ - 34}}{\text{ Js}}\); (allow ECF, eg answers in the range of 6.0 to \(6.8 \times {10^{ - 34}}\))

Do not allow a bald answer of \(6.6 \times {10^{ - 34}}\) as value is known.

\(E = 6.62 \times {10^{ - 34}} \times 9.3 \times {10^{14}}{\text{ }}( = 6.2 \times {10^{ - 19}}{\text{ J}})\);

an answer in the range of 3.4 to 4.0 eV; (allow ECF from (e)(i) and (e)(ii))

or

identifies y intercept as work function;

an answer in the range of 3.6 to 3.8 eV;

straight line of any length parallel to original data;

to left of original data, intercept on \(f\)-axis would be positive;

A number of candidates effectively repeated the question by stating that light had to exceed a threshold frequency to cause photoemission. To gain both marks they were expected to refer to the need for photon energy \((hf)\) to be greater than the work function energy for the metal surface.

To determine the threshold frequency most candidates correctly extended the graph to find the intercept on the f axis. Quite large numbers of candidates misread the scale.

Various methods were used to determine h. Commonly the gradient was measured, but as always candidates seemed to be unaware that they should choose points on their line as far apart as possible. The gradient in eVs was usually correctly converted to Js.

The work function was also determined in a number of different ways. The easiest method was to use the y intercept. Most candidates did well on this question.

Almost all candidates correctly drew a line parallel and to the left of the original line.

A beam of electrons is accelerated from rest through a potential difference of 85 V.

Show that the de Broglie wavelength associated with the electrons in the beam is 0.13 nm.

Electrons with the same kinetic energy as those in (b) are incident on a circular aperture of diameter 1.1 nm.

The electrons are detected beyond the aperture.

The graph shows the variation with angle \(\theta \) of the number n of electrons detected per second after diffraction by the aperture.

Use your answer to (b) opposite to explain how data from the graph support the de Broglie hypothesis.

\({\text{KE of electron}} = 1.4 \times {10^{ - 17}}{\text{ J}}\);

combine \(\lambda  = \frac{h}{p}\) and \({E_{\text{k}}} = \frac{{{p^2}}}{{2m}}\) to get \({\lambda ^2} = \frac{{{h^2}}}{{2m{E_k}}}\);

\({\lambda ^2} = \frac{{{{\left[ {6.6 \times {{10}^{ - 34}}} \right]}^2}}}{{2 \times 9.1 \times {{10}^{ - 31}} \times 1.4 \times {{10}^{ - 17}}}}\);

\(\lambda  = 1.3 \times {10^{ - 10}}{\text{ m}}\)

or

\(v = \sqrt {\frac{{2eV}}{m}} \);

\(p = \sqrt {2meV} \);

\(\lambda  = \left( {\frac{h}{{\sqrt {2meV} }} = } \right){\text{ }}\frac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 85} }}\);

\(\lambda  = 1.3 \times {10^{ - 10}}{\text{ m}}\)

minimum read-off at 0.15 rad; (allow answers in the range of 0.14 rad to 0.16 rad)

\(\lambda  = \frac{{b{\theta _{{\text{min}}}}}}{{1.22}}\);

\( = 1.4 \times {10^{ - 10}}{\text{ m}}\);

same/similar wavelength to (b) (so de Broglie hypothesis supported);

Allow [3 max] if 1.22 missing – answer is then 0.162 nm.

Candidates answered this well and had been well schooled in the demands of this calculation. However a significant number wrote solutions that were completely unintelligible to the examiners.

Candidates were required to use the graph of electron beam intensity against scattering angle to show that the scattering aperture diameter was commensurate with the electron wavelength. About half the candidates could achieve this, but some lost marks through omitting the 1.22 factor in the circular aperture equation.

An alpha particle with initial kinetic energy 32 MeV is directed head-on at a nucleus of gold-197 \(\left( {{}_{79}^{197}{\rm{Au}}} \right)\).

(i) Show that the distance of closest approach of the alpha particle from the centre of the nucleus is about 7×10⁻¹⁵m.

(ii) Estimate the density of a nucleus of \({}_{79}^{197}{\rm{Au}}\) using the answer to (a)(i) as an estimate of the nuclear radius.

The nucleus of \({}_{79}^{197}{\rm{Au}}\) is replaced by a nucleus of the isotope \({}_{79}^{195}{\rm{Au}}\). Suggest the change, if any, to your answers to (a)(i) and (a)(ii).

Distance of closest approach:

Estimate of nuclear density:

An alpha particle is confined within a nucleus of gold. Using the uncertainty principle, estimate the kinetic energy, in MeV, of the alpha particle.

(i)
32 MeV converted using 32×10⁶×1.6×10^–19«=5.12×10^–12J»

\(d =  \ll \frac{{kQq}}{E} = \frac{{8.99 \times {{10}^9} \times 2 \times 79 \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{32 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} =  \gg \frac{{8.99 \times {{10}^9} \times 2 \times 79 \times 1.6 \times {{10}^{ - 19}}}}{{32 \times {{10}^6}}}\)

OR 7.102×10^-15m

«d≈7×10^-15m»

Must see final answer to 2+ SF unless substitution is completely correct with value for k explicit.
Do not allow an approach via \(r = {R_0}{A^{\frac{{\rm{1}}}{{\rm{3}}}}}\).

(ii)
m≈197×1.661×10^-27
OR
3.27×
10^-25kg

\(V = \frac{{4\pi }}{3} \times {\left( {7 \times {{10}^{ - 15}}} \right)^3}\)

OR

1.44×10^-42m^-3

\(\rho  =  \ll \frac{m}{V} = \frac{{3.2722 \times {{10}^{ - 25}}}}{{1.4368 \times {{10}^{ - 42}}}} =  \gg 2.28 \times {10^{17}} \approx 2 \times {10^{17}}{\rm{kg}}{{\rm{m}}^{ - 3}}\)

Allow working in MeV: 1.28×10⁴⁷MeVc^–2m^–3.
Allow ECF from incorrect answers to MP1 or MP2.

Distance of closest approach: charge or number of protons or force of repulsion is the same so distance is the same

Estimate of nuclear density: «\(\rho  \propto \frac{A}{{{{\left( {{A^{\frac{1}{3}}}} \right)}^3}}}\) so» density the same

Δ
x≈7×10^–15 m

\(\Delta p \approx \frac{{6.63 \times {{10}^{ - 34}}}}{{4\pi  \times 7 \times {{10}^{ - 15}}}} \ll  = 7.54 \times {10^{ - 21}}{\rm{Ns}} \gg \)

\(E \approx  \ll \frac{{\Delta {p^2}}}{{2m}} = \frac{{{{\left( {7.54 \times {{10}^{ - 21}}} \right)}^2}}}{{2 \times 6.6 \times {{10}^{ - 27}}}} = 4.3 \times {10^{ - 15}}{\rm{J}} = 26897{\rm{eV}} \gg  \approx 0.027{\rm{MeV}}\)

Accept Δ
x≈3.5×10^–15m or Δ
x≈1.4×10^–14m leading to E≈0.11MeV or 0.0067MeV.
Answer must be in MeV.

(i)     Distinguish between an insulator and a conductor.

(ii)     Outline what is meant by the internal resistance of a cell.

State what is meant by the photoelectric effect.

(i)     Suggest why the work function for caesium is smaller than that of mercury.

(ii)     Ultraviolet radiation of wavelength 210 nm is incident on the surface of mercury. The work function for mercury is 4.5 eV. Determine the maximum kinetic energy of the photoelectrons emitted.

An exact determination of the location of the electron in a hydrogen atom is not possible. Outline how this statement is consistent with the Schrödinger model of the hydrogen atom.

(i)     conductor has free electrons/charges that are free to move within/through it / insulator does not have free electrons/charges that are free to move within/ through it;

electrons act as charge carriers;

when a pd acts across a conductor a current exists when charge (carriers) move;

Do not allow “good/bad conductor/resistor” or reference to conductivity/ resistivity.

(ii)     some of the power/energy delivered by a cell is used/dissipated in driving current through the cell;

power loss can be equated to \({I^2}r\) where r represents the (internal) resistance of the cell; } (symbols must be defined)

resistance of contents of cell; (do not allow “resistance of cell”)

the emission of electrons from a (metal) surface by photons/light/electromagnetic radiation (incident on the surface);

(i)     caesium electrons are less firmly bound / mercury requires more energy to release electron; } (allow reverse argument)

If answer is in terms of threshold frequency, frequency must be linked to energy via \(E = hf\).

(ii)     energy of photon \( = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.1 \times {{10}^{ - 7}}}}{\text{ }}\left( { = 9.5 \times {{10}^{ - 19}}{\text{ (J)}}} \right)\);

convert photon energy to eV, 5.9 (eV) / convert work function to joules \(7.2 \times {10^{ - 19}}{\text{ (J)}}\);

so kinetic energy of electron = (photon energy \( - \) work function =) 1.4 (eV) or \(2.3 \times {10^{ - 19}}{\text{ (J)}}\);

Award [3] for a bald correct answer.

(Schrödinger model suggests) electron is described by wavefunction;

that gives probability of finding electron at a particular place / probability of finding electrons is proportional to square of (wavefunction) amplitude;

so position of electron is uncertain;

(i)     Superficial answers were common. Candidates continue to ignore the mark allocations for questions and therefore misunderstand the number of independent points they should mention in an answer. Here, most said that conductors contain free electrons (or the reverse for insulators) but did not go on to discuss the role of the free electrons in carrying charge or to relate the current to the existence of an electric field across the conductor. Far too many gave answers of the “conductors conduct well” variety that do not score marks.

(ii)     Too often candidates were content to suggest that the internal resistance of a cell is the resistance of the cell contents without discussing the physical implications of this. It was rare to see a consideration of the energy dissipation in the cell or an explanation of the way the power loss is related to a “resistance”.

Many candidates were able to give a complete description of the photoelectric effect.

(i)     Although the majority were able to relate work function to the physics of the electrons in the metal, some could only respond in terms of the minimum frequency required to produce a photocurrent. This did not generally score marks without some supporting remarks.

(ii)     Generally, candidates were able to score at least two marks. Work was marred by power of ten errors and by inabilities to convert between the electrovolt and the joule. A major reason for errors was that candidates often did not begin with a clear statement of the photoelectric equation followed by substitution in an organised way.

Significant numbers scored two out of three marks. There were some good attempts to link the wavefunction idea to the probability ideas of the theory.

Write down the nuclear equation for this decay.

Deduce that the activity of the radium-226 is almost constant during the experiment.

Show that about 3 x 10¹⁵ alpha particles are emitted by the radium-226 in 6 days.

The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.

The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^–5 m³ and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B over the 6 day period. Helium is a monatomic gas.

\(_2^4\alpha \)

OR

\({}_2^4{\text{He}}\)

\({}_{86}^{222}{\text{Rn}}\)

These must be seen on the right-hand side of the equation.

ALTERNATIVE 1

6 days is 5.18 x 10⁵ s

activity after 6 days is \({A_0}{e^{ - 1.4 \times {{10}^{ - 11}} \times 5.8 \times {{10}^5}}} \approx {A_0}\)

OR

A = 0.9999927 A₀ or 0.9999927 \(\lambda \)N₀

OR

states that index of e is so small that \(\frac{A}{{{A_0}}}\) is ≈ 1

OR

A – A₀ ≈ 10^–15 «s^–1»

ALTERNATIVE 2
shows half-life of the order of 10¹¹ s or 5.0 x 10¹⁰ s

converts this to year «1600 y» or days and states half-life much longer than experiment compared to experiment

Award [1 max] if calculations/substitutions have numerical slips but would lead to correct deduction.

eg: failure to convert 6 days to seconds but correct substitution into equation will give MP2.

Allow working in days, but for MP1 must see conversion of \(\lambda \) or half-life to day^–1.

ALTERNATIVE 1 

use of A = \(\lambda \)N₀

conversion to number of molecules = nN_A = 3.7 x 10²⁰

OR

initial activity = 5.2 x 10⁹ «s^–1»

number emitted = (6 x 24 x 3600) x 1.4 x 10^–11 x 3.7 x 10²⁰ or 2.7 x 10¹⁵ alpha particles

ALTERNATIVE 2
use of N = N₀\({e^{ - \lambda t}}\)

N₀ = n x N_A = 3.7 x 10²⁰

alpha particles emitted «= number of atoms disintegrated = N – N₀ =» N₀\(\left( {1 - {e^{ - \lambda  \times 6 \times 24 \times 3600}}} \right)\) or 2.7 x 10¹⁵ alpha particles 

Must see correct substitution or answer to 2+ sf for MP3

alpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule

Do not allow reference to tunnelling.

conversion of temperature to 291 K

p = 4.5 x 10^–9 x 8.31 x «\(\frac{{291}}{{1.3 \times {{10}^{ - 5}}}}\)»

OR

p = 2.7 x 10¹⁵ x 1.3 x 10^–23 x «\(\frac{{291}}{{1.3 \times {{10}^{ - 5}}}}\)»

0.83 or 0.84 «Pa»

Allow ECF for 2.7 x 10¹⁵ from (b)(ii).

Outline how atomic emission spectra provide evidence for the quantization of energy in atoms.

Consider an electron confined in a one-dimensional “box” of length L. The de Broglie waves associated with the electron are standing waves with wavelengths given by \(\frac{{2L}}{n}\), where n=1, 2, 3, …

Show that the energy E_n of the electron is given by

\[En = \frac{{{n^2}{h^2}}}{{8{m_e}{L^2}}}\]

where h is Planck’s constant and m_e is the mass of the electron.

An electron is confined in a “box” of length L=1.0×10^–10m in the n=1 energy level. Its position as measured from one end of the box is (0.5±0.5)×10^–10m. Determine

(i) the momentum of the electron.

(ii) the uncertainty in the momentum.

all particles have an associated wavelength / OWTTE;

wavelength is given by \(\lambda  = \frac{h}{p}\), where h is Planck’s constant and p is momentum;

from de Broglie hypothesis, \({p_n} = \frac{h}{{{\lambda _n}}} = \frac{{nh}}{{2L}}\);
kinetic energy given by \({E_K} = \frac{{{p^2}}}{{2{m_e}}}\);
combined and manipulated to obtain result;

(i) \(\lambda  = \frac{{2L}}{n} = \frac{{2 \times 1.0 \times {{10}^{ - 10}}}}{1} = 2.0 \times {10^{ - 10}}\);
\(p = \frac{h}{\lambda } = \frac{{6.6 \times {{10}^{ - 34}}}}{{2.0 \times {{10}^{ - 10}}}} = 3.3 \times {10^{ - 24}}{\rm{kgm}}{{\rm{s}}^{ - 1}}\);
Award [2] for alternative methods, e.g. calculating energy then momentum.

(ii) use of \(\Delta x\Delta p \ge \frac{h}{{4\pi }}\);
to get \(\Delta p \ge \frac{{6.6 \times {{10}^{ - 34}}}}{{4\pi  \times 0.5 \times {{10}^{ - 10}}}} = 1.1 \times {10^{ - 24}}{\rm{kgm}}{{\rm{s}}^{ - 1}}\);

Monochromatic light is incident on a metal surface and electrons are emitted instantaneously from the surface.

Explain why

(i) the emission of the electrons is instantaneous.

(ii) the energy of the emitted electrons does not depend on the intensity of the incident light.

The wavelength of the incident light in (a) is 420 nm and the work function of the metal is 3.4×10^–19J.

(i) Determine, in joules, the maximum kinetic energy of an emitted electron.

(ii) The metal surface has dimensions of 1.5 mm×2.0 mm. The intensity of the light incident on the surface is 4.5×10^–6 W m^–2. On average, one electron is emitted for every 300 photons that are incident on the surface. Determine the initial electric current leaving the metal surface.

(i) mention of photons;
of quantized energy / energy is hf;
one to one correspondence with electrons and photons / OWTTE;
(so arrival of light causes emission straightaway)

(ii) intensity is a measure of the number of photons not the individual photon energy;

(i) \(E = \frac{{hc}}{{4.2 \times {{10}^{ - 7}}}}\);
4.7×10^–19J;
so maximum energy (4.7×10⁻¹⁹-3.4×10⁻¹⁹=)1.3×10⁻¹⁹J;

(ii) energy arriving per second=3×10^–6×4.5×10^–6(=1.35×10⁻¹¹W);
so arrival rate of photons \( = \frac{{1.35 \times {{10}^{ - 11}}}}{{4.7 \times {{10}^{ - 19}}}}\) } (allow ECF from (b)(i))
(=2.9×10⁷ photons s^–1);
and a current of \(\left( {\frac{{2.9 \times {{10}^7} \times 1.6 \times {{10}^{ - 19}}}}{{300}} = } \right)1.5 \times {10^{ - 14}}{\rm{A}}\);

Describe what is meant by

(i) radioactive decay.

(ii) nuclear fusion.

Tritium is a radioactive nuclide with a half-life of 4500 days. It decays to an isotope of helium.

Determine the time taken for 90% of a sample of tritium to decay.

A nuclide of deuterium \(\left( {{}_1^2{\rm{H}}} \right)\) and a nuclide of tritium \(\left( {{}_1^3{\rm{H}}} \right)\) undergo nuclear fusion. The reaction equation for this process is

\[{}_1^2{\rm{H + }}{}_1^3{\rm{H}} \to {}_2^4{\rm{He + X}}\]

Identify X.

(i) refers to unstable nucleus/isotope / refers to spontaneous/random process;
which emits named radiation (from nucleus) / forms different nucleus/isotope;

(ii) combination of two nuclei / OWTTE; (do not allow “particles” or “atoms”)
to form new nuclide with greater mass/larger nucleus/greater number of nucleons;

\(\lambda  = \frac{{\ln 2}}{{4500}}\left( { = 1.54 \times {{10}^{ - 4}}} \right)\);
\(0.1{N_0} = {N_0}{{\rm{e}}^{ - 1.54 \times {{10}^{ - 4}}t}}\);
1.5×10⁴(d) or 1.3×10⁹(s);
Award [2 max] if answer is time to lose 10% (680 d).
Allow answer to be expressed in any time units.
Award [3] for a bald correct answer.

or

\(\ln 0.1 = \frac{{ - 0.69t}}{{{t_{\frac{1}{2}}}}}\);
t=3.3×4500;
1.5×10⁴(d);
Award [2 max] if answer is time to lose 10% (680 d).
Allow answer to be expressed in any time units.
Award [3] for a bald correct answer.

\({}_0^1n\)/neutron;

Outline why the wave model of light cannot account for the photoelectric effect.

Calculate, in eV, the maximum kinetic energy of the photoelectrons emitted.

The intensity of the light is \({\text{5.1 }}\mu {\text{W}}\,{{\text{m}}^{ - 2}}\). Determine the number of photoelectrons emitted per second for each \({\text{m}}{{\text{m}}^{\text{2}}}\) of the metal surface. Each photon has a 1 in 800 chance of ejecting an electron.

electrons require energy for release;

electrons (are observed) to appear instantaneously;

wave model requires time delay (to build up enough energy);

or

the kinetic energy of the (emitted) electrons depends on frequency (of incident light);

with no electron emission below a threshold frequency;

a wave model suggests emission at all frequencies;

or

Award [2 max] only for this approach.

a maximum electron energy is observed (for a particular wavelength);

a wave model would permit any value so no maximum;

Only allow one route, candidate cannot pick facts from the three alternatives.

(photon) energy \(\frac{{hc}}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}} \times 3.00 \times {{10}^8}}}{{420 \times {{10}^{ - 9}}}}{\text{ }}( = 4.74 \times {10^{ - 19}}{\text{ J}})\);

\({E_{\max }} = (hf - \phi  = ){\text{ }}4.74 \times {10^{ - 19}} - 2.60 \times {10^{ - 19}}\);

1.33 or 1.34 (eV); } (this mark is for correct conversion to eV; allow ECF from incorrect MP1 and MP2)

Award [3] for a bald correct answer.

If no unit given, assume eV and mark accordingly – eg: award [2 max] for a non-conversion.

\(5.1{\text{ }}\mu {\text{W}}\,{{\text{m}}^{ - 2}} = 5.1 \times {10^{ - 12}}{\text{ (J}}{{\text{s}}^{ - 1}}{\text{m}}{{\text{m}}^{ - 2}}{\text{)}}\);

(number of incident photons per \({\text{m}}{{\text{m}}^2}\) per second) = \(\frac{{5.1 \times {{10}^{ - 12}}}}{{4.74 \times {{10}^{ - 19}}}}{\text{ }}( = 1.08 \times {10^7})\);

(number of photoelectrons per \({\text{m}}{{\text{m}}^2}\) per second) = \(\frac{{1.08 \times {{10}^7}}}{{800}}{\text{ }}( = 1.3 \times {10^4})\);

Accept 1.4 \( \times \) \({10^4}\) using rounded energy in (b)(i).

For alternative approaches look for the following in any order:

correct transformation from power/\({{\text{m}}^{\text{2}}}\) to energy/s/\({\text{m}}{{\text{m}}^2}\);

correct use of incoming photon energy; (must see “photon”)

Allow ECF from (b)(i) if identifiable correct insertion of 800 factor and final calculation.

Award [3] for a bald correct answer.

A number of alternative arguments can be used in this question. The most frequent one was the approach via the instantaneous appearance of the electron when radiation of even the lowest intensities is incident. Too many candidates simply quoted some random observations supposing that the examiner would be happy to join up the thinking. However, one route was allowed with arguments that linked the observation quoted with the predictions that would follow from a consideration of the wave model.

Many candidates can now carry out this and similar calculations fluently and confidently.

Again, a large number of correct solutions were seen with many more deficient in one or two aspects of the solution.

Calculate the wavelength of the light.

Electrons emitted from the surface of the photocell have almost no kinetic energy. Explain why this does not contradict the law of conservation of energy.

Radiation of photon energy 5.2 x 10^–19 J is now incident on the photocell. Calculate the maximum velocity of the emitted electrons.

Describe the change in the number of photons per second incident on the surface of the photocell.

State and explain the effect on the maximum photoelectric current as a result of increasing the photon energy in this way.

wavelength = «\(\frac{{hc}}{E} = \frac{{1.99 \times {{10}^{ - 25}}}}{{{\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 19}}}} = \)» 5.7 x 10^–7 «m»

If no unit assume m.

«potential» energy is required to leave surface

Do not allow reference to “binding energy”.
Ignore statements of conservation of energy.

all/most energy given to potential «so none left for kinetic energy»

energy surplus = 1.7 x 10^–19 J

v_max = \(\sqrt {\frac{{2 \times 1.7 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}}  = 6.1 \times {10^5}\) «m s^–1»

Award [1 max] if surplus of 5.2 x 10^–19J used (answer: 1.1 x 10⁶ m s^–1)

«same intensity of radiation so same total energy delivered per square metre per second» 

light has higher photon energy so fewer photons incident per second

Reason is required

1:1 correspondence between photon and electron

so fewer electrons per second

current smaller

Allow ECF from (c)(i)
Allow ECF from MP2 to MP3.

On the diagram, draw the shape of the electric field between the plates.

Calculate the electric field strength between the plates.

Calculate the force on M.

Determine the change in the electric potential energy of M as it moves from the positive to the negative plate.

Construct the nuclear equation for the decay of radium-226.

Radium-226 has a half-life of 1600 years. Determine the time, in years, it takes for the activity of radium-226 to fall to 5% of its original activity.

minimum of three lines equally spaced and distributed, perpendicular to the plates and downwards; edge effect shown; } (condone lines that do not touch plates)

\(4.3 \times {10^5}{\text{ (N}}{{\text{C}}^{ - 1}}{\text{)}}\)

\((F = Eq = ){\text{ }}4.3 \times {10^5} \times 2 \times 1.6 \times {10^{ - 19}}\); (allow ECF from (b)(i))

\(1.4 \times {10^{ - 13}}{\text{ (N)}}\);

Award [2] for a bald correct answer.

\(\Delta {E_{\text{P}}} = q\Delta V\)\(\,\,\,\)or\(\,\,\,\)\(3.2 \times {10^{ - 19}} \times 5.2 \times {10^3}\);

\(1.7 \times {10^{ - 15}}{\text{ (J)}}\);

negative/loss;

\(\left( {_{\;88}^{226}{\text{Ra}} \to _{\;86}^{222}{\text{Rn}} + _2^4{\text{He}} + _0^0\gamma } \right)\)

\(_{\;86}^{222}{\text{Rn}}\)\(\,\,\,\)or\(\,\,\,\)\(_2^4{\text{He}}\);

numbers balance top and bottom on right-hand side;

\(\lambda  = \frac{{\ln 2}}{{1600}} = 4.33 \times {10^{ - 4}}{\text{ (y}}{{\text{r}}^{ - 1}}{\text{)}}\);

\(0.05 = {{\text{e}}^{ - \lambda t}}\);

6900 (years);

Award [3] for a bald correct answer.

Award [2 max] for 2.18 \( \times \) 10¹¹ (s).

Award [1 max] to a candidate who identifies time as about 4.3 half-lives but cannot get further or gives an approximate reasoned answer.

However award [3] if number n of half-lives is calculated from 0.05 = 2^–n (= 4.32 usually from use of log₂ working) and time shown.

Field patterns were often negligently drawn. Lines did not meet both plates, edge effects were ignored, and the (vital) equality of spacing between drawn lines was not considered. Candidates continue to show their inadequacy in responding to questions that demand a careful and accurate diagram.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part. The only common error was the omission of a consideration of the gain or loss of the energy change in part (iii).

As one of the easiest questions on the paper this was predictably well done.

In the past candidates have found calculation involving exponential change difficult. On this occasion, however examiners saw a large number of correct and well explained solutions from candidates.

Show that the energy of photons from the UV lamp is about 10 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

E₁ = –13.6 «eV» E₂ = – \(\frac{{13.6}}{4}\) = –3.4 «eV»

energy of photon is difference E₂ – E₁ = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10^–19 = 7.8 × 10^–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10⁻¹⁹ «J».

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

kinetic energy at collecting plate = 0.9 «eV»

speed = «\(\sqrt {\frac{{2 \times 0.9 \times 1.6 \times {{10}^{ - 19}}}}{{9.11 \times {{10}^{ - 31}}}}} \)» = 5.6 × 10⁵ «ms^–1»

Allow ECF from MP1

[2 marks]

A particular K meson has a quark structure \({\rm{\bar u}}\)s. State the charge, strangeness and baryon number for this meson.

The Feynman diagram shows the changes that occur during beta minus (β^–) decay.

Label the diagram by inserting the four missing particle symbols and the direction of the arrows for the decay particles.

C-14 decay is used to estimate the age of an old dead tree. The activity of C-14 in the dead tree is determined to have fallen to 21% of its original value. C-14 has a half-life of 5700 years.

(i) Explain why the activity of C-14 in the dead tree decreases with time.

(ii) Calculate, in years, the age of the dead tree. Give your answer to an appropriate number of significant figures.

charge: –1«e» or negative or K⁻

strangeness: –1  

baryon number: 0

Negative signs required.
Award [2] for three correct answers, [1 max] for two correct answer and [0] for one correct answer.

correct symbols for both missing quarks

exchange particle and electron labelled W or W^– and e or e^–

Do not allow W⁺ or e⁺ or β⁺. Allow β or β^–.

arrows for both electron and anti-neutrino correct

Allow ECF from previous marking point.

i
number of C-14 atoms/nuclei are decreasing
OR
decreasing activity proportional to number of C-14 atoms/nuclei
OR
A = A₀e^–λt so A decreases as t increases
Do not allow “particles”
Must see reference to atoms or nuclei or an equation, just “C-14 is decreasing” is not enough.

ii
0.21 = (0.5)ⁿ
OR
\(0.21 = {e^{ - \left( {\frac{{\ln 2 \times t}}{{5700}}} \right)}}\)

n = 2.252 half-lives or t =1 2834 «y»
Early rounding to 2.25 gives 12825 y

13000 y rounded correctly to two significant figures:
Both needed; answer must be in year for MP3.
Allow ECF from MP2.
Award [3] for a bald correct answer.

A current is observed on the ammeter when violet light illuminates C. With V held constant the current becomes zero when the violet light is replaced by red light of the same intensity. Explain this observation.

The graph shows the variation of photoelectric current I with potential difference V between C and A when violet light of a particular intensity is used.

The intensity of the light source is increased without changing its wavelength.

(i) Draw, on the axes, a graph to show the variation of I with V for the increased intensity.

(ii) The wavelength of the violet light is 400 nm. Determine, in eV, the work function of caesium.

(iii) V is adjusted to +2.50V. Calculate the maximum kinetic energy of the photoelectrons just before they reach A.

reference to photon
OR
energy = hf or =\(\frac{{hc}}{\lambda }\)

violet photons have greater energy than red photons

when hf > Φ or photon energy> work function then electrons are ejected

frequency of red light < threshold frequency «so no emission»
OR
energy of red light/photon < work function «so no emission»

i
line with same negative intercept «–1.15V»

otherwise above existing line everywhere and of similar shape with clear plateau

Award this marking point even if intercept is wrong.

ii
\(\frac{{hc}}{{\lambda e}} = \) «\(\frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{40 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} = \)» 3.11 «eV»

Intermediate answer is 4.97×10⁻¹⁹ J.

Accept approach using f rather than c/λ

«3.10 − 1.15 =» 1.96 «eV»
Award [2] for a bald correct answer in eV.
Award [1 max] if correct answer is given in J (3.12×10⁻¹⁹ J).

iii

«KE = qVs =» 1.15 «eV»

OR

1.84 x 10⁻¹⁹ «J»

Allow ECF from MP1 to MP2.

adds 2.50 eV = 3.65 eV

OR

5.84 x 10⁻¹⁹ J

Must see units in this question to identify energy unit used.
Award [2] for a bald correct answer that includes units.
Award [1 max] for correct answer without units.

Outline why uranium ore needs to be enriched before it can be used successfully in a nuclear reactor.

(i) One possible waste product of a nuclear reactor is the nuclide caesium-137 \(\left( {{}_{55}^{137}{\rm{Cs}}} \right)\) which decays by the emission of a beta-minus (β– ) particle to form a nuclide of barium (Ba).

State the nuclear reaction for this decay.

(ii) The half-life of caesium-137 is 30 years. Determine the fraction of caesium-137 remaining in the waste after 100 years.

Some waste products in nuclear reactors are good absorbers of neutrons. Suggest why the formation of such waste products requires the removal of the uranium fuel rods well before the uranium is completely used up.

U-238 is much more common than U-235 in ore;
U-235 is more likely to undergo fission / critical amount of U-235 required to ensure fission / OWTTE;
U-238 absorbs neutrons;
U-238 reduces reaction rate in reactor;

(i) \({}_{56}^{137}{\rm{Ba}}\);
\({}_{ - 1}^0{\beta ^ - }\);
anti-neutrino / \(\bar v\);

(ii) \(\lambda  = \left( {\frac{{\ln 2}}{{30}} = } \right)0.0231{\rm{yea}}{{\rm{r}}^{ - 1}}\);
\(\left( {N = {N_0}} \right){{\rm{e}}^{ - 0.0231 \times 100}}\);
0.099 or 9.9%;

proportion of waste builds up in fuel rod as uranium is consumed;
increasing numbers of neutrons will be absorbed;
this reduces the number available to sustain the chain reaction;
build up of waste deforms fuel rod (which can then be difficult to remove);

A particle P moves with simple harmonic motion.

(i) State, with reference to the motion of P, what is meant by simple harmonic motion.

(ii) State the phase difference between the displacement and the velocity of P.

The diagram shows four spectral lines in the visible line emission spectrum of atomic hydrogen.

(i) Outline how such a spectrum may be obtained in the laboratory.

(ii) Explain how such spectra give evidence for the existence of discrete atomic energy levels.

The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression

\({E_n} =  - \frac{{13.6}}{{{n^2}}}\) where n=1, 2, 3 ..........

The visible lines in the spectrum correspond to electron transitions that end at n=2.

(i) Calculate the energy of the level corresponding to n=2.

(ii) Show that the spectral line of wavelength λ=485nm is the result of an electron transition from n=4.

The alpha particles and gamma rays produced in radioactive decay have discrete energy spectra. This suggests that nuclei also possess discrete energy levels. However, beta particles produced in radioactive decay have continuous energy spectra. Describe how the existence of the antineutrino accounts for the continuous nature of beta spectra.

(i) the acceleration (of a particle/P) is (directly) proportional to displacement;
and is directed towards equilibrium/in the opposite direction to displacement;
Do not accept “directed towards the centre”.

(ii) \(\frac{\pi }{2}\)/90°/quarter of a period;

(i) light from a hydrogen discharge tube/hot hydrogen gas/ hydrogen tube with potential difference across it;
is passed onto a prism/diffraction grating;
and then is observed on a screen/through a telescope;
Accept good labelled diagram for explanation of any marking point.

(ii) each wavelength corresponds to the energy of the photon emitted;
when an electron makes a transition from a higher to lower energy level;
since only discrete wavelengths/finite number of wavelengths are present, then only discrete energy levels are present / OWTTE;

(i) −3.40 eV;
Award [0] for omitted negative sign.

(ii) energy difference between levels\( = \frac{{hc}}{{\lambda e}} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.85 \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}}}}\);
=2.55eV;
\(\left[ {3.40 - 2.55} \right] = 0.85 = \frac{{13.6}}{{{n^2}}}\) to give n²=16;
n=4;
Award [3] for reversed argument.

the total emitted energy is shared between the electron and the antineutrino;
the energy/velocity can be shared/distributed in an infinite number of ways / OWTTE;

State and explain the nature of the particle labelled X.

Outline how these experiments are carried out.

Outline why the particles must be accelerated to high energies in scattering experiments.

State and explain one example of a scientific analogy.

Determine the radius of the magnesium-24 nucleus.

Plot the position of magnesium-24 on the graph.

Draw a line on the graph, to show the variation of nuclear radius with nucleon number.

«electron» neutrino

it has a lepton number of 1 «as lepton number is conserved»

it has a charge of zero/is neutral «as charge is conserved»
OR
it has a baryon number of 0 «as baryon number is conserved»

Do not allow antineutrino

Do not credit answers referring to energy

«high energy particles incident on» thin sample

detect angle/position of deflected particles

reference to interference/diffraction/minimum/maximum/numbers of particles

Allow “foil” instead of thin

λ \( \propto \frac{1}{{\sqrt E }}\) OR λ \( \propto \frac{1}{E}\)

so high energy gives small λ

to match the small nuclear size

Alternative 2

E = hf/energy is proportional to frequency

frequency is inversely proportional to wavelength/c = fλ

to match the small nuclear size

Alternative 3

higher energy means closer approach to nucleus

to overcome the repulsive force from the nucleus

so greater precision in measurement of the size of the nucleus

Accept inversely proportional

Only allow marks awarded from one alternative

two analogous situations stated

one element of the analogy equated to an element of physics

eg: moving away from Earth is like climbing a hill where the contours correspond to the equipotentials

Atoms in an ideal gas behave like pool balls

The forces between them only act during collisions

R = 2.7 x 10^–15 x \({2^{\frac{1}{3}}}\)

3.4 – 3.5 x 10^–15 «m»

Allow use of the Fermi radius from the data booklet

correctly plotted

Allow ECF from (d)(i)

single smooth curve passing through both points with decreasing gradient

through origin

Light is incident on a metal surface A. A potential difference is applied between A and an electrode B. Photoelectrons arrive at B and the resulting current is measured by a sensitive ammeter. (Note: the complete electrical circuit is not shown.)

(i) The frequency of the light is reduced until the current measured by the ammeter falls to zero. Explain how Einstein’s photoelectric theory accounts for this observation.

(ii) A different metal surface is used so that a current is again measured. Outline the effect on the photoelectric current when the intensity of the light is doubled and the frequency remains constant.

A photon of energy 6.6×10^–19J is incident upon a clean sodium surface. The work function of sodium is 3.7×10^–19J. The photon causes an electron to be emitted from the surface with the maximum possible kinetic energy. The position of this electron is measured with an uncertainty of 5.0×10^–9m.

Calculate the

(i) momentum of the electron.

(ii) uncertainty in the momentum of the electron.

(i) light consists of photons/quanta/packets of energy;
(each) photon has energy E=hf / photon energy depends on frequency;
a single photon interacts with a single electron giving up all its energy;
a certain amount of energy is required to eject an electron from the metal;
if photon energy is less than this energy/work function/frequency below threshold, no electrons are emitted;

(ii) increasing the intensity increases the photoelectric current;
photocurrent will change as a different metal has a different work function/threshold frequency;

(i) \({E_K} = \left[ {6.6 - 3.7} \right] \times {10^{ - 19}} = 2.9 \times {10^{ - 19}}\left( {\rm{J}} \right)\);
\({E_K} = \frac{{p2}}{{2m}} \Rightarrow p = \sqrt {2m{E_K}}  = \sqrt {2 \times 9.1\left( 1 \right) \times {{10}^{ - 31}} \times 2.9 \times {{10}^{ - 19}}} \);
p=7.2×10^-25(kg ms^-1); (allow answers in the range of 7.2 to 7.3×10^-25)
Award [3] for a bald correct answer.

(ii) \(\Delta p = \frac{{\frac{h}{{4\pi }}}}{{\Delta x}} = \frac{{\frac{{6.6\left( 3 \right) \times {{10}^{ - 34}}}}{{4\pi }}}}{{5.0 \times {{10}^{ - 9}}}}\);
=1.1×10^-26(kg ms^-1);
Award [2] for a bald correct answer.

State what is meant by a wavefunction.

State the position near which this electron is most likely to be found.

Calculate the momentum of the electron.

The energy, in joules, of the electron in a hydrogen atom, is given by \(E = \frac{{2.18 \times {{10}^{ - 18}}}}{{{n^2}}}\) where n is a positive integer. Calculate the wavelength of the photon emitted in a transition from the first excited state of hydrogen to the ground state.

The electron stays in the first excited state of hydrogen for a time of approximately Δt=1.0×10⁻¹⁰s.

(i) Determine the uncertainty in the energy of the electron in the first excited state.

(ii) Suggest, with reference to your answer to (e)(i), why the photons emitted in transitions from the first excited state of hydrogen to the ground state will, in fact, have a small range of wavelengths.

a function whose (absolute squared) value may be used to calculate the probability of finding a particle near a given position / quantity related to the probability of finding an electron near a given position/at a given position;

middle of the box / (near) 0.5×10⁻¹⁰m;

the de Broglie wavelength is 2.0×10⁻¹⁰m;
\(p = \frac{h}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}}}}{{2.0 \times {{10}^{ - 10}}}} = 3.3 \times {10^{ - 24}}{\rm{Ns}}\);

difference in energy is
\(\Delta E\left( { =  - \frac{{2.18 \times {{10}^{ - 18}}}}{{{2^2}}} + \frac{{2.18 \times {{10}^{ - 18}}}}{{{1^2}}}} \right) = 1.635 \times {10^{ - 18}}{\rm{J}}\);
\(\lambda  = \frac{{hc}}{{\Delta E}}\);
\(\lambda  = \left( {\frac{{6.63 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{1.635 \times {{10}^{ - 18}}}}} \right) = 1.22 \times {10^{ - 7}}{\rm{m}}\);

(i) attempt at using the energy – time uncertainty relation;
\(\Delta E\left( { = \frac{h}{{4\pi \Delta t}} = \frac{{6.63 \times {{10}^{ - 34}}}}{{4\pi  \times 1.0 \times {{10}^{ - 10}}}}} \right) = 5.3 \times {10^{ - 25}}{\rm{J}}\);

(ii) the wavelength of the photons is determined by the difference in energy between the two levels;
and that energy difference is not well defined/definite/not always the same (because of the uncertainty principle);

Explain how each observation provides support for the particle theory but not the wave theory of light.

Determine a value for Planck’s constant.

State what is meant by the work function of a metal.

Calculate the work function of barium in eV.

The experiment is repeated with a metal surface of cadmium, which has a greater work function. Draw a second line on the graph to represent the results of this experiment.

Observation 1:
particle – photon energy is below the work function
OR
E = hf and energy is too small «to emit electrons»
wave – the energy of an em wave is independent of frequency

Observation 2:
particle – a single electron absorbs the energy of a single photon «in an almost instantaneous interaction»
wave – it would take time for the energy to build up to eject the electron

attempt to calculate gradient of graph = «\(\frac{{4.2 \times {{10}^{ - 19}}}}{{6.2 \times {{10}^{14}}}}\)»

\( = 6.8 - 6.9 \times {10^{ - 34}}\) «Js»

Do not allow a bald answer of 6.63 x 10^-34 Js or 6.6 x 10^-34 Js.

ALTERNATIVE 1
minimum energy required to remove an electron «from the metal surface»

ALTERNATIVE 2
energy required to remove the least tightly bound electron «from the metal surface»

ALTERNATIVE 1
reading of y intercept from graph in range 3.8 − 4.2 × 10^–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

ALTERNATIVE 2
reading of x intercept from graph «5.8 − 6.0 × 10¹⁴ Hz» and using hf₀ to get 3.8 − 4.2 × 10^–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

line parallel to existing line
to the right of the existing line

U-235 \(\left( {{}_{92}^{235}{\rm{U}}} \right)\) can undergo alpha decay to form an isotope of thorium (Th).

(i) State the nuclear equation for this decay.

(ii) A sample of rock contains a mass of 5.6 mg of U-235 at the present day. The half-life of U-235 is 7.0\( \times \)10⁸ years. Determine the initial mass of the U-235 if the rock sample was formed 3.9\( \times \)10⁹ years ago.

(i) \({}_{92}^{235}{\rm{U}} \to {}_{90}^{231}{\rm{Th}} + {}_2^4\alpha \); (allow He for \(\alpha \); treat charge indications as neutral)

(ii) \(\lambda  = \frac{{1{\rm{n2}}}}{{7.0 \times {{10}^8}}}\left( { = 9.9 \times {{10}^{ - 10}}{\rm{yea}}{{\rm{r}}^{ - 1}}} \right)\);

\({m_0} = 5.6{{\rm{e}}^{3.9 \times {{10}^9} \times 9.9 \times {{10}^{ - 10}}}}\left( {{\rm{mg}}} \right)\)

Identify the missing information for this decay.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10¹¹ atoms of beryllium-10. The present activity of the sample is 8.0 × 10⁻³ Bq.

Determine, in years, the age of the sample.

The temperature in the laboratory is higher than the temperature of the ice sample. Describe one other energy transfer that occurs between the ice sample and the laboratory.

\(_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}} \to _{{\mkern 1mu} {\mkern 1mu} 5}^{10}{\text{B}} + _{ - 1}^{\,\,\,0}{\text{e}} + {\overline {\text{V}} _{\text{e}}}\)

antineutrino AND charge AND mass number of electron \(_{ - 1}^{\,\,\,0}{\text{e}}\), \(\overline {\text{V}} \)

conservation of mass number AND charge \(_{\,\,5}^{10}{\text{B}}\), \(_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}}\)

Do not accept V.

Accept \({\bar V}\) without subscript e.

[2 marks]

λ «= \(\frac{{\ln 2}}{{1.4 \times {{10}^6}}}\)» = 4.95 × 10^–7 «y^–1»

rearranging of A = λN₀e^–λt to give –λt = ln \(\frac{{8.0 \times {{10}^{-3}} \times 365 \times 24 \times 60 \times 60}}{{4.95 \times {{10}^{-7}} \times 7.6 \times {{10}^{11}}}}\) «= –0.400»

t = \(\frac{{ - 0.400}}{{ - 4.95 \times {{10}^{ - 7}}}} = 8.1 \times {10^5}\) «y»

Allow ECF from MP1

[3 marks]

from the laboratory to the sample

conduction – contact between ice and lab surface.

OR

convection – movement of air currents

Must clearly see direction of energy transfer for MP1.

Must see more than just words “conduction” or “convection” for MP2.

[2 marks]

Three ice cubes at a temperature of 0ºC are dropped into a container of water at a temperature of 22ºC. The mass of each ice cube is 25 g and the mass of the water is 330 g. The ice melts, so that the temperature of the water decreases. The thermal capacity of the container is negligible.

(i) The following data are available.

Specific latent heat of fusion of ice = 3.3×10⁵ J kg^–1
Specific heat capacity of water = 4.2×10³ J kg^–1 K^–1

Calculate the final temperature of the water when all of the ice has melted. Assume that no thermal energy is exchanged between the water and the surroundings.

(ii) Explain how the first law of thermodynamics applies to the water when the ice cubes are dropped into it.

Explain how the pattern demonstrates that electrons have wave properties.

Electrons are accelerated to a speed of 3.6×10⁷ ms⁻¹ by the electric field.

(i) Calculate the de Broglie wavelength of the electrons.

(ii) The cathode and anode are 22 mm apart and the field is uniform.
The potential difference between the cathode and the anode is 3.7 kV.
Show that the acceleration of the electrons is approximately 3×10¹⁶ms⁻².

State what can be deduced about an electron from the amplitude of its associated wavefunction.

An electron reaching the central bright spot on the fluorescent screen has a small uncertainty in its position. Outline what the Heisenberg uncertainty principle is able to predict about another property of this electron.

(i) use of M×4.2×103×∆θ;
ml = 75×10–3×3.3×105 / 24750 J;
recognition that melted ice warms and water cools to common final temperature;
3.4°C;

(ii) work done on water by dropping cubes / negligible work done;
W negative or unchanged;
water gives thermal energy to ice;
Q negative;
water cools to a lower temperature;
∆ U negative / U decreases;

square of amplitude (of wavefunction);
(proportional to) probability of finding an electron (at a particular point);

relates position to momentum (or velocity);
large uncertainty in momentum / most information on momentum is lost;

State what is meant by work function.

The diagram shows part of an experimental arrangement used to investigate the photoelectric effect.

(i) Explain how the maximum kinetic energy of the emitted electrons is determined experimentally.

(ii) On the diagram, draw the power supply and other necessary components needed in order to carry out the experiment in (b)(i).

Using results obtained with the apparatus in (b), the following graph was drawn. The graph shows how the maximum kinetic energy of the photoelectrons varies with the frequency of the incident radiation.

State how the graph can be used to determine

(i) a value for the Planck constant.

(ii) the work function of the material.

(iii) the threshold wavelength of the material.

In an experiment, light at a particular frequency is incident on a surface and electrons are emitted. Explain what happens to the number of electrons emitted per second when the intensity of this light is increased.

minimum (photon) energy needed to eject electrons (from a surface);

(i) apply stopping potential / OWTTE;
maximum kinetic energy =eV;

(ii)

power supply negative connected to detector;
ammeter and voltmeter correctly connected;
Allow voltmeter connected directly across power supply.

(i) Planck constant = gradient;

(ii) work function = (−) intercept on maximum kinetic energy (allow ‘y’) axis;

(iii) \({\lambda _0} = \frac{c}{{{\rm{frequency intercept}}}}\) or \(\frac{{hc}}{{{\rm{max ke intercept}}}}\);

more intense light means more photons per second;
so more electrons are ejected (per second);

Most candidates choosing this option were able to answer this well although a minority forgot to mention that this was the minimum energy needed for a photon to eject an electron from a metal surface.

(i) The technique for measuring the maximum kinetic energy of the emitted electrons was poorly known. Centres would be well advised to use a simulation if they do not have the opportunity to actually perform this experiment with a photocell.

(ii) Given that is such a simple circuit this was very badly answered with many ‘circuits’ not being circuits at all – in essence, ignoring the polarity needed for the cell, this is a simple resistance measuring circuit. Few realised that the detector must have a negative voltage to prevent the electrons from reaching it.

(i) In recognising that the gradient gives a value for the Planck constant, most candidates answered this correctly.

(ii) Again, this was well answered although few candidates mentioned that the work function was the negative of the value of the maximum kinetic energy intercept.

(iii) This was more involved and less well answered. Although many realised that the threshold frequency was the intercept on the frequency axis, few went on to say that the threshold wavelength was the speed of light divided by this value.

Only those candidates approaching this by relating the intensity of light to the number of incident photons per second tended to be successful here. By stating this it was a simple matter or recognising that there is a one to one correspondence between photons and electrons so more intense light inevitably meant more electrons emitted per second. Many wasted time in explaining why emission of electrons meant that the incident light had a frequency higher than the threshold frequency.

For the final thallium nuclide, identify the

(i) nucleon number.

(ii) proton number.

Radon-220 is a radioactive gas. It is released by rocks such as granite. In some parts of the world, houses are built from materials containing granite. Explain why it is unlikely that radon-220 will build up in sufficient quantity to be harmful in these houses.

(i) Calculate, in hour⁻¹, the decay constant of lead-212.

(ii) In a pure sample of lead-212 at one instant, 8.0 × 10⁻³ kg of the lead-212 is present. Calculate the mass of lead-212 that remains after a period of 35 hours.

(iii) A sample of pure radium begins to decay by the series shown in the table. At one instant, a mass of 8.0 × 10⁻³ kg of lead-212 is present in the sample. Suggest why, after 35 hours, there will be a greater mass of lead-212 present in the sample than the value you calculated in (h)(ii).

(i) \(\left( {\lambda  = \frac{{{\rm{In2}}}}{{{{\rm{T}}_{\frac{1}{2}}}}} = \frac{{0.693}}{{10.6}} = } \right)6.5 \times {10^{ - 2}}{\rm{ hou}}{{\rm{r}}^{ - 1}}\)

(ii) use of λ from (h)(i);
correct substitution into N = N0e−λt ;
8.0 to 8.3 × 10–4 kg;

(iii) the rate of decay/activity of polonium/radium;
is greater than the rate of decay/activity of lead;

A 24Ω resistor is made from a conducting wire.

(i) The diameter of the wire is 0.30 mm and the wire has a resistivity of 1.7 × 10^–8 Ω m. Calculate the length of the wire.

(ii) A potential difference of 12V is applied between the ends of the wire. Calculate the acceleration of a free electron in the wire.

(iii) Suggest why the average speed of the free electron does not keep increasing even though it is being accelerated.

An electric circuit consists of a supply connected to a 24Ω resistor in parallel with a variable resistor of resistance R. The supply has an emf of 12V and an internal resistance of 11Ω.

Power supplies deliver maximum power to an external circuit when the resistance of the external circuit equals the internal resistance of the power supply.

(i) Determine the value of R for this circuit at which maximum power is delivered to the external circuit.

(ii) Calculate the reading on the voltmeter for the value of R you determined in (b)(i).

(iii) Calculate the power dissipated in the 24Ω resistor when the maximum power is being delivered to the external circuit.

State what is meant by the wavefunction of an electron.

An electron is confined in a length of 2.0 \( \times \) 10^–10 m.

(i) Determine the uncertainty in the momentum of the electron.

(ii) The electron has a momentum of 2.0 \( \times \) 10^–23Ns. Determine the de Broglie wavelength of the electron.

(iii) On the axes, sketch the variation of the wavefunction \(\Psi \) of the electron in (d)(ii) with distance x. You may assume that \(\Psi \) \( = \) 0 when x \( = \) 0.

(iv) Identify the feature of your graph in (d)(iii) that gives the probability of finding the electron at a particular position and at a particular time.

(iv) amplitude of Ψ/graph;
squared;