This question is about the photoelectric effect and the de Broglie hypothesis.

When photons are incident on a lithium surface photoelectrons are emitted. The work function *φ* of lithium is 2.9 eV.

a.

Define *work function.*

Determine the maximum wavelength of the photons that can cause photoemission.

Calculate the momentum of an electron that has the same de Broglie wavelength as the wavelength of the photons in (b).

**Answer/Explanation**

## Markscheme

a.

minimum energy/work required to remove an electron (from the surface of the substance);

\({f_{\min }} = \frac{{2.9 \times 1.6 \times {{10}^{ – 19}}}}{{6.63 \times {{10}^{ – 34}}}} = \left( {7.0 \times {{10}^{14}}} \right)\left( {{\text{Hz}}} \right)\);

\({\lambda _{\max }} = \left( {\frac{{3.00 \times {{10}^8}}}{{7.0 \times {{10}^{14}}}}} \right) = 4.3 \times {10^{ – 7}}\left( {\text{m}} \right)\);

\(p = \left( {\frac{h}{{{\lambda _{\max }}}} = } \right)\frac{{6.63 \times {{10}^{ – 34}}}}{{4.3 \times {{10}^{ – 7}}}}\);

=1.5×10–27 (kg ms–1);*Allow ECF from (b). *

**or **

\(p = \left( {\frac{\phi }{c} = } \right)\frac{{2.9 \times 1.6 \times {{10}^{ – 19}}}}{{3.00 \times {{10}^8}}}\)

=1.5×10–27 (kg ms–1);*Allow ECF from (b).*

This question is about the photoelectric effect.

In a photoelectric experiment, light of wavelength 450 nm is incident on a sodium surface. The work function for sodium is 2.4 eV.

a.

(i) Calculate, in eV, the maximum kinetic energy of the emitted electrons.

(ii) The number of electrons leaving the sodium surface per second is 2 \( \times \) 10^{15}. Calculate the current leaving the sodium surface.

The wavelength of the light incident on the sodium surface is decreased without changing its intensity. Explain why the number of electrons emitted from the sodium will decrease.

**Answer/Explanation**

## Markscheme

a.

(i) photon energy \( = \left( {\frac{{hc}}{\lambda } = \frac{{6.63 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{450 \times {{10}^{ – 9}}}} = } \right)4.4 \times {10^{ – 19}}\left( {\rm{J}} \right)\);

=2.76(eV) ;

2.76 – 2.4 = 0.36 (eV) ;*A**ward [3] for a bald correct answer.Award [1 max] if the energy of the photon is not converted from Joules to eV (giving EK =-2.4 eV ). *

(ii) 2×1015×1.6×10–19 ;

3×10–4 (A);*Award **[2] **for a bald correct answer.*

light consists of photons;

frequency of photons increases so energy of photons increases;

same intensity of radiation means fewer photons;

fewer photons means fewer (photo)electrons;

(the emitted number of electrons falls)

## Examiners report

a.

In part (a)(i) many correct answers were seen, however there were quite a few mistakes in converting the photon energy to eV. Part (a)(ii) was easy, but some candidates thought that the KE of the electrons needed to be used.

Candidates found (b) difficult as the answer is slightly counter-intuitive. Very few candidates seemed to know that equal intensity of light means equal total photon energy per second. Many assumed it meant equal number of photons per second. The answers were mostly disorganised and did not reflect a logical scientific argument that would be expected at this level.

This question is about quantum physics.

a.

Describe the de Broglie hypothesis.

An electron is accelerated from rest through a potential difference of 5.0 kV.

(i) Calculate the momentum of the electron after acceleration.

(ii) Calculate the wavelength of the electron.

(iii) Determine the energy of a photon that has the same wavelength as the electron in (b)(ii).

The momentum of the electron is known precisely. Deduce that all the information on its position is lost.

With reference to Schrödinger’s model, state the meaning of the amplitude of the wavefunction for the electron.

**Answer/Explanation**

## Markscheme

a.

all particles have an associated wavelength/behave like waves;

with \(\lambda = \frac{h}{p}\) and symbols defined/described using terms;

(i) \(p = \left( {\sqrt {2mE} = \sqrt {2meV} = } \right)\sqrt {2 \times 9.11 \times {{10}^{ – 31}} \times 1.6 \times {{10}^{ – 19}} \times 5.0 \times {{10}^3}} \);

\( = 3.8 \times {10^{ – 23}}\left( {{\rm{Ns}}} \right)\);

**or**

\(v = \left( {\sqrt {\frac{{2eV}}{m}} = } \right)\sqrt {\frac{{2 \times 1.6 \times {{10}^{ – 19}} \times 5.0 \times {{10}^3}}}{{9.11 \times {{10}^{ – 31}}}}} \);

\(p = (mv = )3.8 \times {10^{ – 23}}(Ns)\);

(ii) \(\lambda = \left( {\frac{h}{p} = } \right)\frac{{6.63 \times {{10}^{ – 34}}}}{{3.8 \times {{10}^{ – 23}}}}\);

=1.7×10^{-11}m;*This is a question testing units for this option. Do not award second marking point for an incorrect or missing unit.*

(iii) \(E = \left( {hf = \frac{{hc}}{\lambda } = } \right)\frac{{6.63 \times {{10}^{ – 34}} \times 3.0 \times {{10}^8}}}{{1.7 \times {{10}^{ – 11}}}}\);

E=1.2×10^{-14}(J);

*or*

\(E = (cp = )3.0 \times {10^8} \times 3.8 \times {10^{ – 23}}\);

\(E = 1.2 \times {10^{ – 14}}(J)\);*Allow ECF from (b)(ii).*

reference to the Heisenberg uncertainty principle / \(\Delta x\Delta p \ge \frac{h}{{4\pi }}\);

Δ*p *= 0 implies Δ*x* is large /Δ*x*=∞;

the (square of the) amplitude gives the probability of finding the electron at a given point in space;

This question is about the photoelectric effect.

When light is incident on a clean metal surface, electrons can be emitted through the photoelectric effect.

a.

Outline how the Einstein model is used to explain the photoelectric effect.

State why, although the incident light is monochromatic, the energies of the emitted electrons vary.

Explain why no electrons are emitted if the frequency of the incident light is less than a certain value, no matter how intense the light.

For monochromatic light of wavelength 620 nm a stopping potential of 1.75 V is required. Determine the minimum energy required to emit an electron from the metal surface.

**Answer/Explanation**

## Markscheme

a.

light made of photons of energy \(E = hf\);

electrons are released immediately from the metal;

if electron gains sufficient energy (from a photon);

different electrons may be bound by a different amount of energy to the metal;

insufficient photon energy to eject surface electrons;

greater intensity means more photons but still none with enough energy;

\({E_{\max }} = (1.75 \times 1.60 \times {10^{ – 19}} = ){\text{ }}2.80 \times {10^{ – 19}}{\text{ (J)}}\);

\(\phi = \left( {hf – {E_{\max }} = 6.63 \times {{10}^{ – 34}} \times \frac{{3.00 \times {{10}^8}}}{{620 \times {{10}^{ – 9}}}} – 2.80 \times {{10}^{ – 19}} = } \right){\text{ }}4.1 \times {10^{ – 20}}{\text{ (J)}}\);

## Examiners report

a.

The general idea of the photon explanation of the photoelectric effect was well known, but only a few clearly referred to photon energy transfer. Very few could explain why monochromatic light gives varied electron energies, most referring to various frequencies of light as the reason. The common mistake was the electron gaining different amounts of energy from different frequencies (despite monochromatic being stated in the question). While threshold frequency was well understood the effect of intensity was usually overlooked. The calculation was usually well attempted.

The general idea of the photon explanation of the photoelectric effect was well known, but only a few clearly referred to photon energy transfer. Very few could explain why monochromatic light gives varied electron energies, most referring to various frequencies of light as the reason. The common mistake was the electron gaining different amounts of energy from different frequencies (despite monochromatic being stated in the question). While threshold frequency was well understood the effect of intensity was usually overlooked. The calculation was usually well attempted.

The general idea of the photon explanation of the photoelectric effect was well known, but only a few clearly referred to photon energy transfer. Very few could explain why monochromatic light gives varied electron energies, most referring to various frequencies of light as the reason. The common mistake was the electron gaining different amounts of energy from different frequencies (despite monochromatic being stated in the question). While threshold frequency was well understood the effect of intensity was usually overlooked. The calculation was usually well attempted.

This question is about energy level transitions.

Some of the electron energy levels for a hydrogen atom are shown.

A hydrogen atom is excited to the \( – 1.51{\text{ eV}}\) level.

Monochromatic radiation is incident on gaseous hydrogen. All the hydrogen atoms are in the ground state. Describe what could happen to the radiation and to the hydrogen atoms if the incident photon energy is equal to

a.i.

On the diagram, label using arrows all the possible transitions that might occur as the hydrogen atom returns to the ground state.

State the energy in eV of the maximum wavelength photon emitted as the hydrogen atom returns to the ground state.

10.2 eV.

9.0 eV.

**Answer/Explanation**

## Markscheme

a.i.

only the three correct arrows on diagram;

*(–1.51 to –3.40, –1.15 to –13.6 and –3.40 to –13.6) *

1.89 eV; *(allow ECF from diagram) *

photon is absorbed;

electron (in a hydrogen atom) raised to higher/–3.40 eV/excited state;

no absorption / photon pass through;

## Examiners report

a.i.

The hydrogen atom energy diagram was generally well-answered. The energy of the maximum wavelength was usually confused with the maximum frequency. The description of the photons of different energies were usually answered incompletely, not referring to both the radiation and the hydrogen atoms. A number of candidates referred to hydrogen atoms jumping energy levels, rather than electrons.

The hydrogen atom energy diagram was generally well-answered. The energy of the maximum wavelength was usually confused with the maximum frequency. The description of the photons of different energies were usually answered incompletely, not referring to both the radiation and the hydrogen atoms. A number of candidates referred to hydrogen atoms jumping energy levels, rather than electrons.

The hydrogen atom energy diagram was generally well-answered. The energy of the maximum wavelength was usually confused with the maximum frequency. The description of the photons of different energies were usually answered incompletely, not referring to both the radiation and the hydrogen atoms. A number of candidates referred to hydrogen atoms jumping energy levels, rather than electrons.

This question is about the wave nature of matter.

In 1927 Davisson and Germer tested the de Broglie hypothesis. They directed a beam of electrons onto a nickel crystal as shown in the diagram. The experiment was carried out in a vacuum.

a.

Describe wave-particle duality in relation to the de Broglie hypothesis.

The electrons were accelerated through a potential difference of 54 V. Show that the associated de Broglie wavelength for the electrons is about \(2 \times {10^{ – 10}}{\text{ m}}\).

The electron detector recorded a large number of electrons at a particular scattering angle \(\theta \). Explain why a maximum in the number of scattered electrons is observed at a particular angle.

**Answer/Explanation**

## Markscheme

a.

a particle with momentum has a wavelength ;

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

\(p = \sqrt {2{m_{\text{e}}} \times eV} = \sqrt {2 \times 9.11 \times {{10}^{ – 31}} \times 1.60 \times {{10}^{ – 19}} \times 54} {\text{ }}( = 3.97 \times {10^{ – 24}}{\text{ kg}}\,{\text{m}}\,{{\text{s}}^{ – 1}})\);

\(\lambda = \left( {\frac{h}{{\sqrt {2{m_{\text{e}}} \times eV} }} = \frac{{6.63 \times {{10}^{ – 34}}}}{{3.97 \times {{10}^{ – 24}}}} = } \right){\text{ }}1.7 \times {10^{ – 10}}{\text{ m}}\); }

*(must see 2*+ *significant figures to award this mark)*

electrons scatter off the periodic structure of the nickel lattice;

separation of ions/atoms is such that diffraction occurs (leading to a maximum in the intensity of scattered electrons);

## Examiners report

a.

(a) was usually well answered.

(b)(i) was either well answered or incorrect.

(b)(ii) was very poorly answered.

This question is about wave–particle duality.

A particle of mass \({\text{6.4}} \times {\text{1}}{{\text{0}}^{ – {\text{27}}}}{\text{ kg}}\) and charge \(3.2 \times {10^{ – 19}}{\text{ C}}\) is accelerated from rest through a potential difference of 25 kV.

a.

Describe what is meant by the de Broglie hypothesis.

(i) Calculate the kinetic energy of the particle.

(ii) Determine the de Broglie wavelength of the particle.

**Answer/Explanation**

## Markscheme

a.

all particles have an associated wavelength / *OWTTE*;

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

(i) \({E_{\text{K}}}( = 3.2 \times {10^{ – 19}} \times 25 \times {10^3}) = 8.0 \times {10^{ – 15}}{\text{ (J)}}\)\(\,\,\,\)** or**\(\,\,\,\)50 (keV);

(ii) use of \({E_{\text{K}}} = \frac{{{p^2}}}{{2m}}\)\(\,\,\,\)and\(\,\,\,\)\(p = \frac{h}{\lambda }\)\(\,\,\,\)** or**\(\,\,\,\)use of \({E_{\text{K}}} = \frac{1}{2}m{v^2}\)\(\,\,\,\)and\(\,\,\,\)\(p = mv = \frac{h}{\lambda }\);

\(p = 1.0 \times {10^{ – 20}}{\text{ (Ns)}}\);

\(\lambda = \left( {\frac{h}{p} = } \right){\text{ }}6.6 \times {10^{ – 14}}{\text{ (m)}}\)\(\,\,\,\)** or**\(\,\,\,\)\(6.5 \times {10^{ – 14}}{\text{ (m)}}\);

*Award **[3] **for a bald correct answer.*

## Examiners report

a.

The de Broglie hypothesis was sometimes stated poorly and symbols sometimes not defined.

In (i) the kintetic energy was usually correct, but in (ii) far fewer correct answers were seen due to both algebraic and arithmetic errors. A common mistake was to treat the de Broglie wavelength as electromagnetic.

This question is about atomic energy levels.

a.

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

Outline how the de Broglie hypothesis explains the existence of a **discrete** set of wavefunctions for electrons confined in a box of length* L*.

The diagram below shows the shape of two allowed wavefunctions *ѱ _{A}* and

*ѱ*for an electron confined in a one-dimensional box of length

_{B}*L*.

(i) With reference to the de Broglie hypothesis, suggest which wavefunction corresponds to the larger electron energy.

(ii) Predict and explain which wavefunction indicates a larger probability of finding the electron near the position \(\frac{L}{2}\) in the box.

(iii) On the graph in (c) on page 7, sketch a possible wavefunction for the **lowest** energy state of the electron.

**Answer/Explanation**

## Markscheme

a.

atomic spectra have discrete line structures / only discrete frequencies/wavelengths;

photon energy is related to frequency/wavelength;

photons have discrete energies;

photons arise from electron transitions between energy levels;

which must have discrete values of energy;

de Broglie suggests that electrons/all particles have an associated wavelength; this wave will be a stationary wave which meets the boundary conditions of the box; the stationary wave has wavelength \(\frac{{2L}}{n}\) (where *L* is the length of the box and where *n* is an integer);

(i) wavelength of *ψ _{A}* larger than

*ψ*;

_{B}therefore momentum of

*ψ*larger than

_{B}*ψ*(from de Broglie hypothesis); therefore

_{A}*ψ*has larger energy;

_{B}*Award*

**[1 max]**for a bald correct answer.**or**

*ψ _{B}* has

*n*=3,

*ψ*has

_{A}*n*=2;

E

_{K}∝

*n*

^{2};

so

*ψ*corresponds to the larger energy;

_{B}(ii) *ψ _{A}*=0,

*ψ*≠0 in the middle of the box/at \(\frac{L}{2}\);

_{B}so

*ψ*

_{B}corresponds to the larger probability since probability ∝Ι

*ψ*Ι

^{2};

*Accept ∝ ψ*

^{2}.**or**

the probability (of finding the electron) is related to the amplitude;

amplitude of *ψ _{B}* is greater than amplitude of

*ψ*so

_{A}*ψ*is more likely to be found;

_{B}*Award [1 max] for a bald correct answer.*

(iii)

correct sketch; (accept –*ψ*)*Accept wavefunction with any amplitude.*

## Examiners report

a.

(a) Candidates struggled with this question. Although they demonstrated some familiarity with the idea, they could not clearly describe the connection between atomic structure and the emission spectra, usually discussing electrons without photons. The arguments leading from atomic spectra to energy levels were not logically organised.

There were very few correct answers to (b).

(i) was reasonably well done by many, although many did not refer to the de Broglie hypothesis explicitly and thus relate wavelength to momentum and so to energy.

(ii) was poorly answered. Not many candidates understood the relation between amplitude and probability of locating the particle.

(iii) was well done by most.

This question is about the photoelectric effect.

The diagram shows apparatus used to investigate the photoelectric effect.

a.

When red light is incident on the metallic surface M the microammeter registers a current. Explain how a current is established in this circuit even though nothing joins M to C inside the tube.

The graph shows the variation with voltage *V* of the current *I* in the circuit.

The work function of the metallic surface M is 0.48 eV.

(i) Define *work function*.

(ii) State the maximum kinetic energy of an electron immediately after it has been emitted from M.

(iii) Calculate the energy of a photon incident on M.

(iv) The red light incident on M is now replaced by blue light. The number of photons incident on M per second is the same as in (b).

On the axes opposite, sketch a graph to show the variation with *V* of the current *I*.

**Answer/Explanation**

## Markscheme

a.

the light causes emission of (photo)electrons;

which move (from M) to C;

(i) the (minimum) energy required to eject an electron from the metal;

(ii) 1.42 eV;*Allow answer in Joules.*

(iii) (1.42+0.48 )=1.90 eV;

Allow answer in Joules.

(iv) line starting to the left of where red curve starts;

and saturates at the same value as red;

This question is about quarks and interactions.

a.

Outline how interactions in particle physics are understood in terms of exchange particles.

Determine whether or not strangeness is conserved in this decay.

The total energy of the particle represented by the dotted line is 1.2 GeV more than what is allowed by energy conservation. Determine the time interval from the emission of the particle from the s quark to its conversion into the d \({\rm{\bar d}}\) pair.

The pion is unstable and decays through the weak interaction into a neutrino and an anti-muon.

Draw a Feynman diagram for the decay of the pion, labelling all particles in the diagram.

**Answer/Explanation**

## Markscheme

a.

exchange particles are virtual particles/bosons;

that mediate/carry/transmit the weak/strong/em force between interacting particles / *OWTTE*;*Award first marking point for named bosons also, e.g. photons, W, Z, gluons.*

strangeness in initial state is –1 and zero in the final;

hence it is not conserved;

*Award [0] for unsupported second marking point.*

\(\Delta t \approx \frac{h}{{4\pi \Delta E}} = \frac{{6.63 \times {{10}^{ – 34}}}}{{4\pi \times 1.2 \times {{10}^9} \times 1.6 \times {{10}^{ – 19}}}}\);

\(\Delta t \approx 3 \times {10^{ – 25}}{\rm{s}}\);

diagram as above;

correctly labelled W^{+};

*Allow time to run vertically. Allow particle symbols. Ignore missing or wrong arrow directions.*

This question is about the de Broglie hypothesis.

a.

State the de Broglie hypothesis.

Determine the de Broglie wavelength of a proton that has been accelerated from rest through a potential difference of 1.2 kV.

Explain why a precise knowledge of the de Broglie wavelength of the proton implies that its position cannot be observed.

**Answer/Explanation**

## Markscheme

a.

particles have an associated wavelength;

wavelength=\(\frac{h}{{mv}}\) or \(\frac{h}{{p}}\); *(symbols must be defined)*

\(\lambda = \frac{h}{{\sqrt {2{\rm{meV}}} }}\);

8.3×10^{−13}m;

(Heisenberg suggests that) Δ*p*Δ*x* is a constant * or *\( \ge \frac{h}{{4\pi }}\);

if

*λ*is known then Δ

*p*is zero therefore uncertainty in position Δ

*x*is infinite/very large;

*Award*

**[1 max]**if Δp and Δx not defined.**or**

(the Uncertainty Principle states that) it is impossible to know the position and momentum of a particle at the same time;

if λ is precise then momentum is precise so position is not known;

This question is about the photoelectric effect.

a.

State what is meant by the photoelectric effect.

Light of frequency 8.7×10^{14}Hz is incident on the surface of a metal in a photocell. The surface area of the metal is 9.0×10^{–6}m^{2} and the intensity of the light is 1.1×10^{–3}Wm^{–2}.

(i) Deduce that the maximum possible photoelectric current in the photocell is 2.7 nA.

(ii) The maximum kinetic energy of photoelectrons released from the metal surface is 1.2 eV. Calculate the value of the work function of the metal.

**Answer/Explanation**

## Markscheme

a.

ejection of electron from metal surface following absorption of em radiation/photon;

(i) energy of one photon = 6.67×10^{−34}×8.7×10^{14}(=5.8×10^{−19}J);

number of electrons released from surface per second \( = \frac{{9.0 \times {{10}^{ – 6}} \times 1.1 \times {{10}^{ – 3}}}}{{5.8 \times {{10}^{ – 19}}}}\);

=1.7×10^{10};

current=1.7×10^{10}×1.6×10^{−19};

= 2.7nA

(ii) 2.4eV * or* 3.9×10

^{−19}J;

This question is about the photoelectric effect.

a.

Describe the concept of a photon.

In the photoelectric effect there exists a threshold frequency below which no emission of photoelectrons takes place.

Outline how the

(i) wave theory of light is unable to account for this observation.

(ii) concepts of the photon and work function are able to account for this observation.

Light of wavelength 420 nm is incident on a clean metal surface. The work function of the metal is 2.0 eV.

Determine the

(i) threshold frequency for this metal.

(ii) maximum kinetic energy in eV of the emitted electrons.

**Answer/Explanation**

## Markscheme

a.

light consists of discrete packets/quanta/bundles of energy/particle;

each photon has an energy of *hf* (where *h* is the Planck constant and *f* is the frequency of light);

(i) the energy of a (em) wave depends on amplitude (not frequency);

so increasing the intensity should have resulted in electrons being emitted (at any frequency) / *OWTTE*;

(ii) the work function is the minimum energy required to eject an electron from a metal surface;

if the photon energy (*hf* ) is less than the work function then no emission will take place;

(i) recognizes that work function = h×threshold frequency;

\({f_0} = \left( {\frac{{2.0 \times 1.6 \times {{10}^{ – 19}}}}{{6.6 \times {{10}^{ – 34}}}} = } \right)4.8 \times {10^{14}}{\rm{Hz}}\);

(ii) recognize that maximum KE=*hf-hf*_{0} **or***hf*–*Φ*;

\({f_0} = \left( {\frac{c}{\lambda } = \frac{{3.0 \times {{10}^8}}}{{4.2 \times {{10}^{ – 7}}}} = } \right)7.14 \times {10^{14}}{\rm{Hz}}\);

\(hf\left( {{\rm{eV}}} \right) = \left( {\frac{{6.6 \times {{10}^{ – 34}} \times 7.14 \times {{10}^{14}}}}{{1.6 \times {{10}^{ – 19}}}} = } \right)2.96{\rm{eV}}\);

max KE=(2.96–2.0=)0.96eV;

This question is about the photoelectric effect.

a.

The diagram shows the set up of an experiment designed to verify the Einstein model of the photoelectric effect.

The tungsten electrode is positive.

Explain how the maximum kinetic energy of electrons ejected from the positive electrode is determined.

Light of frequency *f* is shone onto the tungsten electrode in (a). The potential *V*_{s} for which the photoelectric current is zero is recorded for different values of *f*.

(i) Using the axes below, sketch a graph of how you might expect *V*_{s} to vary with *f*.

(ii) State the Einstein photoelectric equation in a form that relates *V*_{s} and *f*. Define, other than the electron charge, any other symbols that you might use.

(iii) Outline how a graph of *V*_{s} against *f* can be used to find the Planck constant and work function of tungsten.

The work function of tungsten is 4.5eV. Show that the de Broglie wavelength of an electron that has this energy is about 0.6nm.

**Answer/Explanation**

## Markscheme

a.

the potential difference is varied (using the potential divider);

until the current registered by the ammeter is zero;

the maximum kinetic energy of the (ejected) electrons is this potential times the electron charge;

(i)

straight line;

with non-zero intercept on *f* axis;

(ii) *V*_{s}*e*=*hf*–*hf*_{0} **or***V*_{s}*e*=*hf*–*φ*;*f*_{0}→the frequency below which no electron emission takes place;

h→the Planck constant;*φ*→the minimum energy required to eject an electron from tungsten;*Award [2 max] if the equation is not given.*

(iii) *Planck constant:* slope/gradient of graph×*e*;*work function:* extrapolation to intercept on *V*_{s} axis and *φ*=*V*_{s}-intercept×*e* / when *V*_{s}=0, *φ*=*hf* so intercept gives *f* when *V*_{s}=0 and *φ*=*h* (*f*-intercept);

use of \(p = \frac{h}{\lambda }\) and \({E_k} = \frac{{{p^2}}}{{2m}}\);

\(\lambda = \frac{h}{{\sqrt {2{E_{\rm{k}}}m} }}\);

\( = \frac{{6.6 \times {{10}^{ – 34}}}}{{2 \times 4.5 \times 1.6 \times {{10}^{ – 19}} \times 9.1 \times {{10}^{ – 31}}}} = 5.765 \times {10^{ – 10}}{\rm{m}}\);

≈0.6nm

This question is about the photoelectric effect.

a.

Monochromatic light of different frequencies is incident on a metal surface placed in a vacuum. As the frequency is increased a value is reached at which electrons are emitted from the surface. Below this frequency, no matter how intense the light, no electrons are emitted. Outline how the

(i) wave theory of light is unable to account for these observations.

(ii) Einstein model of the photoelectric effect is able to account for these observations.

The graph shows how the maximum kinetic energy *E*_{K} of the ejected electrons in (a) varies with the frequency *f* of the incident light.

Use the graph to determine the

(i) Planck constant.

(ii) work function of the metal.

Show that electrons of energy 0.50 eV have a de Broglie wavelength of about 1.7×10^{–9}m.

<

**Answer/Explanation**

h2 style=”margin-top: 1em;”>Markscheme

a.

(i) electrons in the metal require a minimum amount of energy to be ejected from the metal;

(according to wave theory) the energy of a wave is dependent on intensity and not frequency;

so given enough time to absorb energy electron emission should take place at any frequency no matter what the intensity / *OWTTE*;

(ii) photons have energy *hf*/proportional to frequency (of the light);

an electron may be ejected if this energy is equal to or greater than a threshold value/work function;

the intensity determines the rate of release of photoelectrons, but not their energy;

(i) recognize that slope of graph\( = \frac{h}{e}\) **or***h* (in eV s);

evidence of finding slope eg. \(\frac{{0.5}}{{[6.8 – 5.6] \times {{10}^{14}}}} = 4.17 \times {10^{ – 15}}\); } (accept values in the range of 4.0 to 4.2×10^{-15})

\(h = 1.6 \times {10^{ – 19}} \times 4.17 \times {10^{ – 15}} = 6.7 \times {10^{ – 34}}\left( {{\rm{Js}}} \right)\); } (accept values in the range of 6.4 and 6.7×10^{-34}(Js))*Award [0] for an unsupported correct answer.*

(ii) threshold frequency=5.6×10^{14}(Hz);

work function (*hf*_{0})=6.63×10^{-34}×5.6×10^{-14}=3.7×10^{-19}(J) * or* 2.3(eV);

*If necessary award*

**[2]**for use of ECF value of h from (b)(i).*Award*

**[2]**for use of any data point and W=hf-Ek giving an answer of*3.7(±0.1)×10*

^{-19}(J).*Award*

**[2]**for a bald correct answer.use \(p = \frac{h}{\lambda }\) and \({E_{\rm{K}}} = \frac{{{p^2}}}{{2m}}\) to show that \(\lambda = \frac{h}{{\sqrt {2m{E_{\rm{K}}}} }}\); *(allow equivalent working)*

electron kinetic energy=0.5×1.6×10^{-19}(J) * or* 8.0×10

^{-20}(J);

\(\lambda = \left( {\frac{{6.63 \times {{10}^{ – 34}}}}{{\sqrt {2 \times 9.11 \times {{10}^{ – 31}} \times 8.0 \times {{10}^{ – 20}}} }} = } \right)1.74 \times {10^{ – 9}}\left( {\rm{m}} \right)\); }

*(must see to three significant figures or better)*

(≈1.7×10

^{-9}m)

*Award marks for evidence of valid working, as the answer is given in the question.*

This question is about the photoelectric effect.

In an experiment to investigate the photoelectric effect, light of frequency ƒ is incident on the metal surface A, shown in the diagram below. A potential difference is applied between A and B. The photoelectric current is measured by a sensitive ammeter. (Note: the complete electrical circuit is not shown.)

When the frequency of the light is reduced to a certain value, the current measured by the ammeter becomes zero. Explain how Einstein’s photoelectric theory accounts for this observation.

**Answer/Explanation**

## Markscheme

light consists of photons/quanta/packets of energy;

each photon has energy *E*=*hf* / photon energy depends on frequency;

a certain amount of energy is required to eject an electron from the metal;

if photon energy is less than this energy, no electrons are emitted;

## Examiners report

This question is about wave–particle duality.

In the photoelectric effect, electrons are not emitted from the surface of a metal if the frequency of the incident light is below a certain value called the threshold frequency.

Light of frequency \(1.0 \times {10^{15}}{\text{ Hz}}\) is incident on the surface of a metal. The work function of the metal is \(3.2 \times {10^{ – 19}}{\text{ J}}\).

a.

(i) Explain, with reference to the Einstein model of the photoelectric effect, the existence of the threshold frequency.

(ii) State, with reference to your answer in (a)(i), the reason why the threshold frequency is different for different metals.

(i) Show that the maximum kinetic energy of the emitted electrons is \(3.4 \times {10^{ – 19}}{\text{ J}}\).

(ii) Determine the de Broglie wavelength of the electrons in (b)(i).

**Answer/Explanation**

## Markscheme

a.

(i) light consists of photons/quanta;

a certain minimum amount of energy (the work function) is required to remove an electron from the metal;

if the photon energy is below this energy/work function no electrons will be emitted;

the energy of the photons is proportional to the frequency / \(E = hf\) (with terms defined);

*If work function is mentioned it must be defined to award **[4]**.*

(ii) different metals need a different amount of minimum energy for electrons to be removed;

*Accept answers in terms of work function if defined either here or in (a)(i).*

(i) \(K{E_{\max }} = hf – \phi \);

\( = 6.6 \times {10^{ – 34}} \times 1.0 \times {10^{15}} – 3.2 \times {10^{ – 19}}\);

\( = 3.4 \times {10^{ – 19}}{\text{ J}}\)

(ii) use of \(E = \frac{{{p^2}}}{{2m}}\) and \(p = \frac{h}{\lambda }\)\(\,\,\,\,\,\)** or**\(\,\,\,\,\,\)use of \(v = \sqrt {\frac{{2E}}{m}} \) and \(p = mv = \frac{h}{\lambda }\);

to give \(\lambda = \frac{h}{{\sqrt {2mE} }}\);

\(\lambda = 8.4 \times {10^{ – 10}}{\text{ m}}\);

## Examiners report

a.

In (a) (i) many candidates showed clearly that they understood the concept of the Einstein model and the existence of a threshold frequency. However, a significant minority of candidates had very little understanding of the topic. Those who answered (i) well had no problem in answering part (ii) correctly.

The problem on maximum kinetic energy was often done well but the standard calculation of the de Broglie wavelength was often done poorly with many candidates unable to make a start.