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Questions on collisions of electrons with atoms
An atom hit by a free electron takes exactly one energy-level gap, which is excitation, or enough to free an electron, which is ionisation. Anything less leaves the collision elastic. A fluorescent tube uses both processes, then turns the resulting ultraviolet into visible light with a phosphor coating. Level diagrams are labelled in electronvolts.
19 original questions · 55 marks · the collisions of electrons with atoms notes · Quantum phenomena
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Distinguish between excitation and ionisation of an atom by a colliding electron.
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Excitation lifts an electron to a higher energy level within the atom (1); ionisation gives the electron enough energy to be removed from the atom completely (1).Calculate the energy, in J, equivalent to 3.4 eV (e = 1.60 × 10−19 C).
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3.4 eV = 3.4 × 1.60 × 10−19 (1)
= 5.44 × 10−19 J (1)Calculate the energy, in eV, equivalent to 8.0 × 10−19 J.
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(8.0 × 10−19)/(1.60 × 10−19) (1)
= 5.0 eV (1)Define the ionisation energy of an atom.
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The minimum energy needed (1) to remove an electron completely from an atom in its ground state (1).An electron is accelerated from rest through a potential difference of 250 V. State its kinetic energy, in electronvolts.
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250 eV: an electron gains one electronvolt for each volt of accelerating pd (1).An electron and a proton are each accelerated from rest through a potential difference of 100 V. State and explain how their final kinetic energies compare.
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They are equal, both 100 eV (1). The energy transferred is QV, and the two charges have the same magnitude (1).Define the electronvolt, and express 12.1 eV in joules.
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One electronvolt is the energy gained by an electron accelerated through a potential difference of one volt (1). 12.1 eV = 12.1 × 1.60 × 10−19 = 1.94 × 10−18 J (1).An electron is accelerated from rest through a potential difference of 5.0 V. Calculate its kinetic energy in joules and its final speed (me = 9.11 × 10−31 kg).
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Ek = eV = 1.60 × 10−19 × 5.0 (1)
Ek = 8.00 × 10−19 J (1)
v = √(2Ek/m) = 1.33 × 106 m s−1 (1)An atom has a first excitation energy of 4.9 eV. An electron with a kinetic energy of 6.0 eV collides with it and excites it. State whether excitation is possible and calculate the kinetic energy the electron has left afterwards, in joules.
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Yes: 6.0 eV exceeds 4.9 eV (1), so excitation occurs (1)
The electron loses 4.9 eV, keeping 6.0 − 4.9 = 1.1 eV (1)
= 1.76 × 10−19 J (1)A hydrogen atom has an ionisation energy of 13.6 eV. Calculate the minimum speed an electron must have to ionise it (me = 9.11 × 10−31 kg).
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Ek = 13.6 eV = 2.18 × 10−18 J (1)
v = √(2Ek/m) (1)
v = 2.19 × 106 m s−1 (1)A mercury atom has an ionisation energy of 10.4 eV. An electron with 12.0 eV of kinetic energy collides with a mercury atom and ionises it. The electron freed from the atom moves away with 0.4 eV of kinetic energy. Calculate the kinetic energy, in J, of the incident electron after the collision (e = 1.60 × 10−19 C).
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Energy is conserved: the incident electron keeps what is left after paying the ionisation energy and the freed electron's share (1). 12.0 − 10.4 − 0.4 = 1.2 eV (1) = 1.92 × 10−19 J (1).In a fluorescent tube, de-exciting mercury atoms emit ultraviolet photons of energy 4.9 eV. The phosphor coating absorbs each ultraviolet photon and emits a photon of visible light of wavelength 546 nm. Calculate the energy per photon, in eV, retained by the coating, and state what becomes of this energy (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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Visible photon: E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(546 × 10−9) = 3.64 × 10−19 J = 2.28 eV (1). Energy retained = 4.9 − 2.28 = 2.6 eV per photon (1). It is transferred to internal (thermal) energy of the coating (1).The neon atoms in an indicator lamp have a first excitation energy of 16.6 eV. Electrons are accelerated from rest between the lamp's electrodes. (a) State the minimum pd across the lamp for excitation of neon to be possible, giving a reason. (b) Calculate the kinetic energy, in J, of an electron accelerated through this pd (e = 1.60 × 10−19 C).
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(a) 16.6 V, because an electron gains one electronvolt per volt of pd, so it then carries exactly 16.6 eV (1). (b) Ek = 16.6 × 1.60 × 10−19 (1) = 2.66 × 10−18 J (1).In a sodium street lamp the sodium atoms have a first excitation energy of 2.1 eV. An electron accelerated through 5.0 V makes two successive exciting collisions with different sodium atoms. Assume the electron gains no further energy from the field between the collisions. Calculate its kinetic energy, in J, after the second collision, and state what happens when it next collides with a sodium atom (e = 1.60 × 10−19 C).
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Initial kinetic energy = 5.0 eV (1). After two excitations: 5.0 − (2 × 2.1) = 0.8 eV (1) = 1.3 × 10−19 J (1). This is less than 2.1 eV, so the next collision is elastic: the atom absorbs nothing and the electron keeps its kinetic energy (1).Explain how excitation and ionisation lead to the emission of light in a fluorescent tube.
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Accelerated electrons collide with mercury atoms, exciting or ionising them (1). As the excited atoms de-excite they emit ultraviolet photons (1). These strike the phosphor coating on the tube (1), which absorbs them and re-emits the energy as visible light (1).An atom has excitation energies of 4.9 eV and 6.7 eV. It is struck by an electron accelerated through 12 V. State which excitations are possible, and calculate the wavelength of the photon emitted when the atom de-excites from the 4.9 eV level (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1).
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The electron carries 12 eV, so both excitations, 4.9 eV and 6.7 eV, are possible (1)
For the 4.9 eV transition: f = E/h = (4.9 × 1.60 × 10−19)/(6.63 × 10−34) (1)
f = 1.18 × 1015 Hz (1)
λ = c/f = 254 nm (1)Explain why an electron with less than an atom's first excitation energy passes through the gas without transferring energy to the atoms.
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The atom can only accept energy in the exact amounts that lift an electron between its energy levels (1). An electron with less than the first excitation energy cannot supply a whole step, so no energy is absorbed (1) and the collision is elastic: the electron continues with its kinetic energy unchanged (1).An electron travelling at 1.3 × 106 m s−1 approaches a mercury atom. The first excitation energy of mercury is 4.9 eV. Deduce whether this electron can excite the atom (me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C).
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Ek = ½mv2 = 0.5 × 9.11 × 10−31 × (1.3 × 106)2 (1) = 7.70 × 10−19 J (1) = 4.8 eV (1). This is less than the 4.9 eV needed, so the electron cannot excite the atom: the collision is elastic (1).The table gives the first excitation energy of four gases: helium 19.8 eV, neon 16.6 eV, mercury 4.9 eV, sodium 2.1 eV. A manufacturer builds a discharge lamp in which electrons are accelerated from rest through a pd of 6.5 V. Deduce which of these gases could emit light in this lamp.
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Each electron gains a maximum of 6.5 eV from the pd (1). A gas can be excited, and so emit photons on de-excitation, only if the electron's energy reaches its first excitation energy (1). Mercury (4.9 eV) and sodium (2.1 eV) can be excited, so either could emit light (1). Helium (19.8 eV) and neon (16.6 eV) cannot: those collisions are elastic and no photons are emitted (1).
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