Question 7
7
The photoelectric effect may be represented by the equation photon energy = work function energy + maximum kinetic energy of electron.
structured6 marks
Question 7(a)
7(a)
State what is meant by work function energy.
Easystructured1 marks
Answer
minimum energy to remove an electron from the metal/surface
Question 7(b)
7(b)
The variation with frequency f of the maximum kinetic energy \(E_{\mathrm{K}}\) of photoelectrons emitted from the surface of sodium metal is shown in Fig. 7.1. Use the gradient of the graph of Fig. 7.1 to determine a value for the Planck constant h. Show your working. h= Js
Mediumstructured2 marks
Answer
gradient \(=4.17 \times 10^{-15}\) (allow \(4.1 \rightarrow 4.3\) ) \(h=4.15 \times 10^{-15} \times 1.6 \times 10^{-19}\) or h=4.1 to \(4.3 \times 10^{-15} \mathrm{eVs}\)
Question 7(c)
7(c)
The sodium metal in (b) has a work function energy of 2.4 eV . The sodium is replaced by calcium which has a work function energy of 2.9 eV . On Fig. 7.1, draw a line to show the variation with frequency f of the maximum kinetic energy \(E_{\mathrm{K}}\) of photoelectrons emitted from the surface of calcium.
Mediumstructured3 marks
Answer
graph: straight line parallel to given line with intercept at any higher frequency
Question 9
9
For a particular metal surface, it is observed that there is a minimum frequency of light below which photoelectric emission does not occur. This observation provides evidence for a particulate nature of electromagnetic radiation.
structured8 marks
Question 9(a)
9(a)
State three further observations from photoelectric emission that provide evidence for a particulate nature of electromagnetic radiation. 1. 2. 3.
Mediumstructured3 marks
Answer
e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity (any three separate statements, one mark each, maximum 3)
Question 9(b)
9(b)
Some data for the variation with frequency f of the maximum kinetic energy \(E_{\mathrm{MAX}}\) of electrons emitted from a metal surface are shown in Fig. 9.1.
structured5 marks
Question 9(b)(i)
9(b)(i)
Explain why emitted electrons may have kinetic energy less than the maximum at any particular frequency.
Mediumstructured2 marks
Answer
(photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1
Question 9(b)(ii)
9(b)(ii)
Use Fig.9.1 to determine 1. the threshold frequency, 2. the work function energy, in eV, of the metal surface. work function energy = eV
Mediumstructured3 marks
Answer
1. threshold frequency \(=5.8 \times 10^{14} \mathrm{~Hz}\) A1 [1] 2. \(\Phi=h f_{0}\) C1 or chooses point on line and substitutes values \(E_{\text {max }}, f\) and h into equation with the units of the h f term converted from J to eV
Question 7
7
Some data for the work function energy \(\Phi\) and the threshold frequency \(f_{0}\) of some metal surfaces are given in Fig. 7.1.
structured9 marks
Question 7(a)
7(a)
4 marks
Question 7(a)(i)
7(a)(i)
State what is meant by the threshold frequency.
Easystructured2 marks
Answer
lowest frequency of e.m. radiation giving rise to emission of electrons (from the surface)
Question 7(a)(ii)
7(a)(ii)
Calculate the threshold frequency for platinum. threshold frequency = Hz
Mediumstructured2 marks
Answer
E=h f threshold frequency \(=\left(9.0 \times 10^{-19}\right) /\left(6.63 \times 10^{-34}\right)\)
Question 7(b)
7(b)
Electromagnetic radiation having a continuous spectrum of wavelengths between 300 nm and 600 nm is incident, in turn, on each of the metals listed in Fig. 7.1. Determine which metals, if any, will give rise to the emission of electrons.
Mediumstructured2 marks
Answer
either \(300 \mathrm{~nm} \equiv 10 \times 10^{15} \mathrm{~Hz}\) (and \(600 \mathrm{~nm} \equiv 5.0 \times 10^{14} \mathrm{~Hz}\) ) or \(\quad 300 \mathrm{~nm} \equiv 6.6 \times 10^{-19} \mathrm{~J}\left(\right.\) and \(\left.600 \mathrm{~nm} \equiv 3.3 \times 10^{-19} \mathrm{~J}\right)\) or zinc \(\lambda_{0}=340 \mathrm{~nm}\), platinum \(\lambda_{0}=220 \mathrm{~nm}\) (and sodium \(\lambda_{0}=520 \mathrm{~nm}\) ) M1 emission from sodium and zinc
Question 7(c)
7(c)
When light of a particular intensity and frequency is incident on a metal surface, electrons are emitted. State and explain the effect, if any, on the rate of emission of electrons from this surface for light of the same intensity and higher frequency.
Mediumstructured3 marks
Answer
each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1
Question 7
7
Some data for the work function energy \(\Phi\) and the threshold frequency \(f_{0}\) of some metal surfaces are given in Fig. 7.1.
structured9 marks
Question 7(a)
7(a)
4 marks
Question 7(a)(i)
7(a)(i)
State what is meant by the threshold frequency.
Easystructured2 marks
Answer
lowest frequency of e.m. radiation giving rise to emission of electrons (from the surface)
Question 7(a)(ii)
7(a)(ii)
Calculate the threshold frequency for platinum. threshold frequency = Hz
Mediumstructured2 marks
Question 7(b)
7(b)
Electromagnetic radiation having a continuous spectrum of wavelengths between 300 nm and 600 nm is incident, in turn, on each of the metals listed in Fig. 7.1. Determine which metals, if any, will give rise to the emission of electrons.
Mediumstructured2 marks
Answer
either \(300 \mathrm{~nm} \equiv 10 \times 10^{15} \mathrm{~Hz}\) (and \(600 \mathrm{~nm} \equiv 5.0 \times 10^{14} \mathrm{~Hz}\) ) or \(\quad 300 \mathrm{~nm} \equiv 6.6 \times 10^{-19} \mathrm{~J}\left(\right.\) and \(\left.600 \mathrm{~nm} \equiv 3.3 \times 10^{-19} \mathrm{~J}\right)\) or zinc \(\lambda_{0}=340 \mathrm{~nm}\), platinum \(\lambda_{0}=220 \mathrm{~nm}\) (and sodium \(\lambda_{0}=520 \mathrm{~nm}\) ) M1 emission from sodium and zinc
Question 7(c)
7(c)
When light of a particular intensity and frequency is incident on a metal surface, electrons are emitted. State and explain the effect, if any, on the rate of emission of electrons from this surface for light of the same intensity and higher frequency.
Mediumstructured3 marks
Answer
each photon has larger energy fewer photons per unit time fewer electrons emitted per unit time M1 M1 A1
Question 7
7
Experiments are conducted to investigate the photoelectric effect.
structured11 marks
Question 7(a)
7(a)
It is found that, on exposure of a metal surface to light, either electrons are emitted immediately or they are not emitted at all. Suggest why this observation does not support a wave theory of light.
Mediumstructured3 marks
Answer
for a wave, electron can 'collect' energy continuously B1 for a wave, electron will always be emitted / electron will be emitted at all frequencies..... M1 after a sufficiently long delay A1
Question 7(b)
7(b)
Data for the wavelength \(\lambda\) of the radiation incident on the metal surface and the maximum kinetic energy \(E_{\mathrm{K}}\) of the emitted electrons are shown in Fig. 7.1.
structured4 marks
Question 7(b)(i)
7(b)(i)
Without any calculation, suggest why no value is given for \(E_{\mathrm{K}}\) for radiation of wavelength 650 nm .
Mediumstructured1 marks
Answer
either wavelength is longer than threshold wavelength or frequency is below the threshold frequency or photon energy is less than work function B1
Question 7(b)(ii)
7(b)(ii)
Use data from Fig. 7.1 to determine the work function energy of the surface. work function energy = J
Mediumstructured3 marks
Answer
\(h c / \lambda=\phi+E_{\mathrm{MAX}}\) C1 \(\left(6.63 \times 10^{-34} \times 3.0 \times 10^{8}\right) /\left(240 \times 10^{-9}\right)=\phi+4.44 \times 10^{-19} \quad \mathrm{C} 1\) \(\phi=3.8 \times 10^{-19} \mathrm{~J}\) (allow \(3.9 \times 10^{-19} \mathrm{~J}\) ) A1 [3]
Question 7(c)
7(c)
Radiation of wavelength 240 nm gives rise to a maximum photoelectric current I. The intensity of the incident radiation is maintained constant and the wavelength is now reduced. State and explain the effect of this change on
structured4 marks
Question 7(c)(i)
7(c)(i)
the maximum kinetic energy of the photoelectrons,
Mediumstructured2 marks
Answer
photon energy larger M1 so (maximum) kinetic energy is larger A1
Question 7(c)(ii)
7(c)(ii)
the maximum photoelectric current I.
Mediumstructured2 marks
Answer
fewer photons (per unit time) M1 so (maximum) current is smaller A1 [2]
Question 8
8
6 marks
Question 8(b)
8(b)
It has been observed that, where photoelectric emission of electrons takes place, there is negligible time delay between illumination of the surface and emission of an electron. State three other pieces of evidence provided by the photoelectric effect for the particulate nature of electromagnetic radiation. 1. 2.
Mediumstructured3 marks
Answer
threshold frequency rate of emission is proportional to intensity max. kinetic energy of electron dependent on frequency max. kinetic energy independent of intensity (any three, 1 each, max 3) B3 [3]
Question 8(c)
8(c)
The work function of a metal surface is 3.5 eV . Light of wavelength 450 nm is incident on the surface. Determine whether electrons will be emitted, by the photoelectric effect, from the surface. Please turn over for Section B. Answer all the questions in the spaces provided.
Mediumstructured3 marks
Answer
either \(E=h c / \lambda\) or \(h c / \lambda=e V\) C1 \(\lambda=450 \mathrm{~nm}\) to give work function of 3.5 eV energy \(=4.4 \times 10^{-19}\) or 2.8 eV to give \(\lambda=355 \mathrm{~nm}\) M1 2.8 eV<3.5 eV so no emission \(355 \mathrm{~nm}<450 \mathrm{~nm}\) so no A1 or work function =3.5 eV threshold frequency \(=8.45 \times 10^{14} \mathrm{~Hz}\) C1 \(450 \mathrm{~nm}=6.67 \times 10^{14} \mathrm{~Hz}\) M1 \(6.67 \times 10^{14} \mathrm{~Hz}<8.45 \times 10^{14} \mathrm{~Hz}\) A1
Question 7
7
An explanation of the photoelectric effect includes the terms photon energy and work function energy.
structured6 marks
Question 7(a)
7(a)
Explain what is meant by
structured1 marks
Question 7(a)(ii)
7(a)(ii)
work function energy.
Easystructured1 marks
Answer
minimum energy to cause emission of an electron (from surface)
Question 7(b)
7(b)
In an experiment to investigate the photoelectric effect, a student measures the wavelength \(\lambda\) of the light incident on a metal surface and the maximum kinetic energy \(E_{\max }\) of the emitted electrons. The variation with \(E_{\max }\) of \(\frac{1}{\lambda}\) is shown in Fig. 7.1.
structured5 marks
Question 7(b)(i)
7(b)(i)
The work function energy of the metal surface is \(\Phi\). State an equation, in terms of \(\lambda, \Phi\) and \(E_{\text {max }}\), to represent conservation of energy for the photoelectric effect. Explain any other symbols you use.
Mediumstructured2 marks
Answer
\(h c / \lambda=\Phi+E_{\max }\)
Question 7(b)(ii)
7(b)(ii)
Use your answer in (i) and Fig. 7.1 to determine 1. the work function energy \(\Phi\) of the metal surface, 2. a value for the Planck constant. Planck constant = Js
Hardstructured3 marks
Answer
1. either when \(1 / \lambda=0, \Phi=-E_{\max }\) or evidence of use of x-axis intercept from graph or chooses point close to the line and substitutes values of \(1 / \lambda\) and \(\Phi=4.0 \times 10^{-19} \mathrm{~J}\left(\right.\) allow \(\left.\pm 0.2 \times 10^{-19} \mathrm{~J}\right)\) 2. either gradient of graph is 1 / h c gradient \(=4.80 \times 10^{24} \rightarrow 5.06 \times 10^{24}\) \(h=1 /\left(\right.\) gradient \(\left.\times 3.0 \times 10^{8}\right)\) or chooses point close to the line and substitutes values of \(1 / \lambda\) and values of \(1 / \lambda\) and \(E_{\text {max }}\) are correct within half a square (Allow full credit for the correct use of any appropriate method) (Do not allow 'circular' calculations in part 2 that lead to the same value of Planck constant that was substituted in part 1)
Question 8
8
9 marks
Question 8(a)
8(a)
Describe two phenomena associated with the photoelectric effect that cannot be explained using a wave theory of light. 1 2
Mediumstructured2 marks
Answer
two from: - there is a frequency below which electrons are not ejected - maximum energy of electron depends on frequency - maximum energy of electrons does not depend on intensity - emission of electrons is instantaneous B2
Question 8(b)
8(b)
The maximum energy \(E_{\text {max }}\) of electrons emitted from a metal surface when illuminated by light of wavelength \(\lambda\) is given by the expression where h is the Planck constant and c is the speed of light.
structured5 marks
Question 8(b)(i)
8(b)(i)
Identify the symbol \(\lambda_{0}\).
Easystructured1 marks
Answer
( \(\lambda_{0}\) is the) threshold wavelength or wavelength corresponding to threshold frequency or maximum wavelength for emission (of electrons) B1
Question 8(b)(ii)
8(b)(ii)
The variation with \(\frac{1}{\lambda}\) of \(E_{\max }\) for the metal surface is shown in Fig. 8.1. Use Fig. 8.1 to determine the magnitude of \(\lambda_{0}\).
Mediumstructured2 marks
Answer
\[ \text { intercept }=1 / \lambda_{0}=2.2 \times 10^{6} \mathrm{~m}^{-1} \] \[ \lambda_{0}=4.5 \times 10^{-7} \mathrm{~m} \] A1
Question 8(b)(iii)
8(b)(iii)
Use the gradient of Fig. 8.1 to determine a value for the Planck constant h. Show your working.
Hardstructured2 marks
Answer
gradient \(=2.0 \times 10^{-25}\) or correct substitution into gradient formula B1 gradient =h c C1 \[ \begin{aligned} h=\left(2.0 \times 10^{-25}\right) /\left(3.0 \times 10^{8}\right) h=6.7 \times 10^{-34} \mathrm{~J} \mathrm{~s} \end{aligned} \] A1
Question 8(c)
8(c)
The metal surface in (b) becomes oxidised. Photoelectric emission is still observed but the work function energy is increased. On Fig. 8.1, draw a line to show the variation with \(\frac{1}{\lambda}\) of \(E_{\max }\) for the oxidised surface.
Mediumstructured2 marks
Answer
line with same gradient as printed line B1 straight line with positive gradient, intercept at greater than \(2.2 \times 10^{6}\) when candidate's line extrapolated B1