Electromagnetic radiation and quantum phenomenaAQA A-Level Physics: Topic test

20 questions, 54 marks

AQA A-Level Physics

Electromagnetic radiation and quantum phenomena topic test

Total 54 marks

Name

Class

Date

1
A clean sodium surface has a work function of 2.3 eV. It is illuminated with monochromatic light of frequency 7.5×10¹⁴ Hz. Use h = 6.63×10⁻³⁴ J s and e = 1.60×10⁻¹⁹ C.
(a)
What is the threshold frequency of sodium?
[1 mark]
A7.5×10¹⁴ Hz
B3.5×10¹⁴ Hz
C5.6×10¹⁴ Hz
D1.3×10¹⁵ Hz
(b)
The intensity of the light is doubled and its frequency is unchanged. What happens to the photoelectrons?
[1 mark]
AThe number emitted per second doubles and their maximum kinetic energy is unchanged.
BThe number emitted per second is unchanged and their maximum kinetic energy doubles.
CThe number emitted per second and their maximum kinetic energy both double.
DNo photoelectrons are emitted, because the intensity is too high.
(c)
Calculate the maximum kinetic energy, in eV, of the photoelectrons emitted from the sodium surface.
[2 marks]
Total for question 1: 4 marks
2
In a gas-discharge tube, electrons are accelerated from rest through a potential difference of 21.0 V and collide with helium atoms. The lowest excitation energy of helium is 19.8 eV and its ionisation energy is 24.6 eV. Use e = 1.60×10⁻¹⁹ C and an electron mass of 9.11×10⁻³¹ kg.
(a)
What is the kinetic energy, in joules, gained by an electron accelerated through 21.0 V?
[1 mark]
A21.0 J
B1.31×10²⁰ J
C3.36×10⁻²⁰ J
D3.36×10⁻¹⁸ J
(b)
Which statement describes what can happen when one of these electrons collides with a helium atom in its ground state?
[1 mark]
AIt can ionise the atom but not excite it.
BIt can excite the atom but not ionise it.
CIt can both excite and ionise the atom.
DIt can neither excite nor ionise the atom.
(c)
An electron excites a helium atom, losing exactly 19.8 eV of its kinetic energy. Calculate the speed of the electron after the collision.
[2 marks]
Total for question 2: 4 marks
3
Some of the energy levels of an atom are: n = 1 at −8.6 eV, n = 2 at −4.9 eV, n = 3 at −3.1 eV and n = 4 at −2.0 eV. Use h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹ and e = 1.60×10⁻¹⁹ C.
(a)
Calculate the frequency and the wavelength of the photon emitted when an electron falls from n = 2 to n = 1.
[3 marks]
(b)
A sample of these atoms has been excited so that all the electrons are in the n = 3 level. State how many different wavelengths can be emitted as the electrons return to the ground state, and calculate the longest of these wavelengths.
[4 marks]
Total for question 3: 7 marks
4
A zinc plate has a work function of 4.30 eV. It is illuminated with ultraviolet light of wavelength 254 nm from a lamp. Use h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹, e = 1.60×10⁻¹⁹ C and m = 9.11×10⁻³¹ kg.
(a)
Calculate the maximum kinetic energy of the photoelectrons in joules and the stopping potential. Explain why visible light of wavelength 400 nm cannot release photoelectrons from the plate, however intense it is.
[6 marks]
(b)
Calculate the de Broglie wavelength of the fastest photoelectrons emitted by 254 nm light. Explain how the photoelectric effect and electron diffraction together illustrate wave–particle duality.
[6 marks]
Total for question 4: 12 marks
5
A beam of neutrons, each with a speed of 2.2×10³ m s⁻¹, is directed at a crystal. Use h = 6.63×10⁻³⁴ J s and a neutron mass of 1.67×10⁻²⁷ kg.
(a)
What is the de Broglie wavelength of the neutrons?
[1 mark]
A1.8×10⁻¹⁰ m
B3.3×10⁻⁷ m
C1.6×10⁻¹³ m
D5.5×10⁹ m
(b)
The neutron speed is doubled. What happens to the angle at which the first diffraction maximum is observed?
[1 mark]
AIt increases, because the wavelength increases.
BIt is unchanged, because the crystal spacing is unchanged.
CIt decreases, because the wavelength decreases.
DIt decreases, because the wavelength increases.
(c)
A tennis ball of mass 0.057 kg travels at 20 m s⁻¹. Calculate its de Broglie wavelength and explain why the ball does not show observable diffraction.
[2 marks]
Total for question 5: 4 marks
6
A helium–neon laser emits red light of wavelength 633 nm when electrons in neon atoms fall between two energy levels. Use h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹ and e = 1.60×10⁻¹⁹ C.
(a)
What is the energy of one photon of the laser light?
[1 mark]
A1.05×10⁻²⁷ J
B3.14×10⁻¹⁹ J
C4.74×10¹⁴ J
D1.96 J
(b)
Why does a gas of excited atoms emit a line spectrum?
[1 mark]
AElectrons in atoms can have any energy, so photons of all energies are emitted.
BPhotons are emitted only when the atoms are ionised.
CAll the electrons fall to the lowest energy level at the same time.
DElectrons in atoms have only certain discrete energies, so photons are emitted only with energies equal to the differences between levels.
(c)
The transition ends on a level of energy −2.60 eV. Show that the photon energy is 1.96 eV and calculate the energy of the level from which the electron falls.
[2 marks]
Total for question 6: 4 marks
7
Light of wavelength 420 nm is incident on a caesium surface that has a work function of 2.10 eV. Use h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹, e = 1.60×10⁻¹⁹ C and m = 9.11×10⁻³¹ kg.
(a)
Calculate the maximum kinetic energy of the photoelectrons, in joules.
[3 marks]
(b)
Calculate the threshold frequency of caesium and the stopping potential for the 420 nm light. State, with a reason, how the stopping potential changes if the intensity of the light is halved.
[4 marks]
Total for question 7: 7 marks
8
In an experiment, electrons are accelerated through a potential difference V in a tube containing mercury vapour at low pressure. As V is increased, the current of electrons reaching a collecting plate falls sharply when V reaches 4.9 V, and ultraviolet radiation is detected coming from the vapour. Use h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹, e = 1.60×10⁻¹⁹ C and m = 9.11×10⁻³¹ kg.
(a)
Explain these observations and calculate the wavelength of the ultraviolet radiation.
[6 marks]
(b)
The ionisation energy of mercury is 10.4 eV. Calculate the speed of an electron accelerated through 4.9 V. Explain what can happen when an electron is accelerated through 11.0 V. Explain why a mercury atom in its ground state cannot absorb a photon of wavelength 300 nm, given that its lowest excitation energy is 4.9 eV.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).