Specification: AQA A-level Physics (7408) 3.12.1.2 Turning points option | Paper: 3 Section B | Data and formulae booklet: ½mv² = eV is given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Thermionic emission: the release of electrons from the surface of a heated metal (the cathode). Heating gives some conduction electrons enough kinetic energy to escape the metal surface.
Accelerating pd: the electrons are accelerated towards a positive anode. The work done on each electron, eV, equals the kinetic energy it gains, so ½mv² = eV (assuming the electrons leave the cathode with negligible speed).
Electron volt: the energy gained by an electron when it is accelerated through a potential difference of 1 volt. 1 eV = 1.60 × 10−19 J.
The equation
kinetic energy gained = charge × accelerating pd ½mv² = eV
| Quantity | Symbol | Unit |
|---|---|---|
| mass of the electron | m (mₑ) | kilograms (kg) |
| speed of the electron at the anode | v | metres per second (m s⁻¹) |
| charge of the electron | e | coulombs (C) |
| accelerating potential difference (anode pd) | V | volts (V) |
Rearranged: v = √(2eV ÷ m) | V = mv² ÷ 2e | Convert kV to V (× 10³) and keV to J (× 1.60 × 10−16) before you substitute, and take the square root last. This non-relativistic form is reliable for anode pds up to about 10 kV.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An electron is accelerated from rest through a potential difference of 2000 V. Calculate the kinetic energy it gains, to 3 significant figures. (e = 1.60 × 10−19 C)
Show solution
F: Ek = eV
I: Ek = 1.60 × 10−19 × 2000
F: Ek is already the subject
A: 3.20 × 10−16 J (3 s.f.), which is 2000 eV
Q2 (Core) In an electron gun, electrons are accelerated from rest through a potential difference of 1500 V. Calculate the speed of the electrons as they reach the anode, to 3 significant figures. Ignore relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
F: ½mv² = eV
I: ½ × 9.11 × 10−31 × v² = 1.60 × 10−19 × 1500
F: v = √(2eV ÷ m) = √(2 × 1.60 × 10−19 × 1500 ÷ 9.11 × 10−31) = √(5.268… × 1014)
A: 2.295… × 107 m s⁻¹ = 2.30 × 107 m s⁻¹ (3 s.f.)
Watch out: forgetting the square root. v² = 5.27 × 1014 is not a speed; a value above 3 × 108 m s⁻¹ should tell you something has gone wrong.
Q3 (Core) Electrons leave the anode of an electron gun at 3.0 × 107 m s⁻¹. Calculate the accelerating potential difference, in kV, to 3 significant figures. Ignore relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
F: ½mv² = eV
I: ½ × 9.11 × 10−31 × (3.0 × 107)² = 1.60 × 10−19 × V
F: V = mv² ÷ 2e = (9.11 × 10−31 × 9.0 × 1014) ÷ (2 × 1.60 × 10−19)
A: 2562.1… V = 2.56 kV (3 s.f.)
Q4 (Core) Each electron in a cathode ray tube gains 4.5 keV of kinetic energy in the electron gun. Calculate the speed of the electrons, to 3 significant figures. Ignore relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: 4.5 keV = 4500 eV = 4500 × 1.60 × 10−19 = 7.2 × 10−16 J
F: Ek = ½mv²
I: 7.2 × 10−16 = ½ × 9.11 × 10−31 × v²
F: v = √(2 × 7.2 × 10−16 ÷ 9.11 × 10−31)
A: 3.975… × 107 m s⁻¹ = 3.98 × 107 m s⁻¹ (3 s.f.)
Watch out: substituting 4.5 or 4500 straight into ½mv². Energies in eV or keV must be changed to joules first.
Q5 (Exam) The anode pd of an electron gun is increased from 1.2 kV to 4.8 kV. (a) Calculate the speed of the electrons at each pd, to 3 significant figures. (b) State the factor by which the speed increases and explain why it is not 4. Ignore relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: 1.2 kV = 1200 V; 4.8 kV = 4800 V
F: v = √(2eV ÷ m)
I: v₁ = √(2 × 1.60 × 10−19 × 1200 ÷ 9.11 × 10−31); v₂ = √(2 × 1.60 × 10−19 × 4800 ÷ 9.11 × 10−31)
F: v is already the subject
A: (a) 2.05 × 107 m s⁻¹ and 4.11 × 107 m s⁻¹ (3 s.f.)
(b) the speed increases by a factor of 2: the kinetic energy is proportional to V, so v ∝ √V, and √4 = 2
Q6 (Exam) An electron gun has an anode pd of 2.5 kV and produces a beam current of 0.40 mA. Calculate (a) the number of electrons reaching the anode each second, and (b) the total kinetic energy carried by the beam each second, each to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
Convert first: 0.40 mA = 4.0 × 10−4 A; 2.5 kV = 2500 V
Step 1 – (a) electrons per second
N = I ÷ e = 4.0 × 10−4 ÷ 1.60 × 10−19 = 2.5 × 1015 electrons per second (2 s.f.)
Step 2 – (b) energy of each electron
F: Ek = eV
I: Ek = 1.60 × 10−19 × 2500 = 4.0 × 10−16 J
F: energy per second = N × Ek = 2.5 × 1015 × 4.0 × 10−16
A: 1.0 J each second (1.0 W, 2 s.f.), the same as P = IV = 4.0 × 10−4 × 2500
Q7 (Exam) Electrons are accelerated from rest through 1800 V and then travel at constant speed through a field-free region to a screen 0.25 m away. Calculate the time taken to reach the screen, to 2 significant figures. Ignore relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – speed at the anode
F: ½mv² = eV
I: ½ × 9.11 × 10−31 × v² = 1.60 × 10−19 × 1800
F: v = √(2 × 1.60 × 10−19 × 1800 ÷ 9.11 × 10−31) = 2.514… × 107 m s⁻¹
Step 2 – time at constant speed
t = distance ÷ speed = 0.25 ÷ 2.514… × 107
A: 9.94… × 10−9 s = 9.9 ns (2 s.f.)
Q8 (Exam) A student uses ½mv² = eV to find the speed of electrons accelerated through 150 kV in an electron microscope. (a) Calculate the speed she obtains, to 2 significant figures, and express it as a fraction of c. (b) Explain why her answer is not reliable. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg, c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 150 kV = 1.50 × 105 V
F: v = √(2eV ÷ m)
I: v = √(2 × 1.60 × 10−19 × 1.50 × 105 ÷ 9.11 × 10−31)
F: v is already the subject
A: (a) 2.295… × 108 m s⁻¹ = 2.3 × 108 m s⁻¹ (2 s.f.), about 0.77c
(b) at this speed relativistic effects are significant: the electron’s mass increases, so ½mv² with the rest mass is not valid. A relativistic calculation gives about 0.63c, so the true speed is lower than her value.
Watch out: only use ½mv² = eV for anode pds up to about 10 kV. If your speed comes out as a large fraction of c, say that relativistic effects cannot be ignored.
Q9 (Stretch) Electrons are accelerated from rest through 2.0 kV and then enter, horizontally and midway between them, two parallel horizontal plates 5.0 cm long and 2.0 cm apart, with a pd of 150 V between the plates. Calculate the vertical deflection of the beam as it leaves the plates, to 2 significant figures. Ignore gravity and relativistic effects. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: 2.0 kV = 2000 V; 5.0 cm = 0.050 m; 2.0 cm = 0.020 m
Step 1 – horizontal speed from the electron gun
v = √(2eV ÷ m) = √(2 × 1.60 × 10−19 × 2000 ÷ 9.11 × 10−31) = 2.650… × 107 m s⁻¹
Step 2 – time between the plates
t = 0.050 ÷ 2.650… × 107 = 1.886… × 10−9 s
Step 3 – vertical acceleration
a = F ÷ m = eVp ÷ md = (1.60 × 10−19 × 150) ÷ (9.11 × 10−31 × 0.020) = 1.317… × 1015 m s⁻²
Step 4 – vertical deflection (uniform acceleration from zero vertical velocity)
y = ½at² = ½ × 1.317… × 1015 × (1.886… × 10−9)²
A: 2.34… × 10−3 m = 2.3 mm (2 s.f.), less than the 10 mm to a plate, so the beam leaves the plates
Watch out: the plates only change the vertical motion. The horizontal speed stays at the value from ½mv² = eV, just like a horizontally launched projectile.
Q10 (Stretch) A student measures the speed v of electrons for several anode pds V. A graph of v² (y-axis) against V (x-axis) is a straight line through the origin passing through the point (3000 V, 1.05 × 1015 m² s⁻²). (a) Explain why the graph is a straight line through the origin and state what its gradient represents. (b) Use the graph to find the specific charge e/m of the electron, to 3 significant figures. (c) Calculate the percentage difference from the value of e/mₑ calculated from the data below, to 1 significant figure. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – (a) the shape of the graph
½mv² = eV rearranges to v² = (2e ÷ m) × V, which has the form y = kx, so v² ∝ V and the gradient = 2e ÷ m
Step 2 – (b) the gradient
gradient = 1.05 × 1015 ÷ 3000 = 3.50 × 1011 m² s⁻² V⁻¹
F: gradient = 2e ÷ m
I: 3.50 × 1011 = 2 × (e ÷ m)
F: e ÷ m = 3.50 × 1011 ÷ 2
A: 1.75 × 1011 C kg⁻¹ (3 s.f.)
Step 3 – (c) comparison
from the data: e ÷ mₑ = 1.60 × 10−19 ÷ 9.11 × 10−31 = 1.756… × 1011 C kg⁻¹
percentage difference = (1.756… − 1.75) ÷ 1.756… × 100 = 0.35…%
A: 0.4% (1 s.f.)
Watch out: the gradient is 2e/m, not e/m. Forgetting to halve it doubles the specific charge.
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