Specification: AQA A-level Physics (7408) 3.2.1.1 | Paper: 1 | Data and formulae booklet: constants e, mₑ, mₚ and mₙ (and the proton and electron specific charges) given; learn specific charge = Q ÷ m
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📌 KEY DEFINITIONS – learn them word for word
Specific charge: the ratio of a particle’s charge to its mass, specific charge = Q ÷ m, measured in C kg⁻¹.
Nucleus and nuclide notation: a nucleus AZX contains Z protons (proton number) and A − Z neutrons, where A is the nucleon number. Its charge is +Ze and its mass is approximately Zmₚ + (A − Z)mₙ.
Ion: an atom that has gained or lost one or more electrons, so it has an overall charge. Removing n electrons gives a charge of +ne; adding n electrons gives −ne.
The equation
specific charge = charge ÷ mass specific charge = Q ÷ m
| Quantity | Symbol | Unit |
|---|---|---|
| specific charge | Q ÷ m | coulombs per kilogram (C kg⁻¹) |
| charge | Q | coulombs (C) |
| mass | m | kilograms (kg) |
Rearranged: Q = specific charge × m | m = Q ÷ specific charge | Count the particles first: charge = (number of protons − number of electrons) × e, and mass = (protons × mₚ) + (neutrons × mₙ) + (electrons × mₑ). Use the magnitude of the charge unless the question asks for the sign.
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) Calculate the specific charge of a proton, to 3 significant figures. (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg)
Show solution
F: specific charge = Q ÷ m
I: specific charge = 1.60 × 10−19 ÷ 1.673 × 10−27
F: specific charge is already the subject
A: 9.563… × 107 = 9.56 × 107 C kg⁻¹ (3 s.f.)
The data booklet lists 9.58 × 107 C kg⁻¹, which is calculated from more precise values of e and mₚ.
Q2 (Core) Calculate the magnitude of the specific charge of an electron, to 3 significant figures, and state how many times larger it is than the specific charge of a proton (9.56 × 107 C kg⁻¹), to 3 significant figures. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
F: specific charge = Q ÷ m
I: specific charge = 1.60 × 10−19 ÷ 9.11 × 10−31
F: specific charge is already the subject
A: 1.756… × 1011 = 1.76 × 1011 C kg⁻¹ (3 s.f.)
ratio = 1.756… × 1011 ÷ 9.563… × 107 = 1836.4… = 1840 times larger (3 s.f.), because the charges are equal in size but the proton is about 1840 times more massive
Q3 (Core) An ion has a mass of 3.32 × 10−26 kg and a specific charge of 4.82 × 106 C kg⁻¹. Calculate its charge, and deduce how many electrons were removed from the atom to form it. (e = 1.60 × 10−19 C)
Show solution
F: specific charge = Q ÷ m
I: 4.82 × 106 = Q ÷ 3.32 × 10−26
F: Q = 4.82 × 106 × 3.32 × 10−26
A: 1.600… × 10−19 C = 1.60 × 10−19 C (3 s.f.)
number of electrons removed = 1.60 × 10−19 ÷ 1.60 × 10−19 = 1 (a singly charged positive ion)
Q4 (Core) A sodium ion, Na⁺, has a mass of 3.82 × 10−23 g. Calculate its specific charge, to 3 significant figures. (e = 1.60 × 10−19 C)
Show solution
Convert first: 1 g = 10−3 kg, so m = 3.82 × 10−23 × 10−3 = 3.82 × 10−26 kg; one electron removed, so Q = +1.60 × 10−19 C
F: specific charge = Q ÷ m
I: specific charge = 1.60 × 10−19 ÷ 3.82 × 10−26
F: specific charge is already the subject
A: 4.188… × 106 = 4.19 × 106 C kg⁻¹ (3 s.f.)
Watch out: leaving the mass in grams. That gives an answer 1000 times too small.
Q5 (Exam) A carbon atom is fully ionised to form a ¹²C⁶⁺ ion (a bare ¹²₆C nucleus). Calculate the specific charge of the ion, to 3 significant figures. (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg, mₙ = 1.675 × 10−27 kg, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – count the particles
6 protons, 12 − 6 = 6 neutrons, 0 electrons
Q = 6 × 1.60 × 10−19 = 9.60 × 10−19 C
m = 6 × 1.673 × 10−27 + 6 × 1.675 × 10−27 = 2.0088 × 10−26 kg
Step 2 – specific charge
F: specific charge = Q ÷ m
I: specific charge = 9.60 × 10−19 ÷ 2.0088 × 10−26
F: specific charge is already the subject
A: 4.778… × 107 = 4.78 × 107 C kg⁻¹ (3 s.f.)
Watch out: using the charge of one proton (e) instead of 6e, or using 12 × mₚ for the mass. Count protons and neutrons separately.
Q6 (Exam) (a) Calculate the specific charge of an alpha particle (⁴₂He nucleus), to 3 significant figures. (b) Compare it with your answer for the ¹²C⁶⁺ ion (4.78 × 107 C kg⁻¹) and with the proton (9.56 × 107 C kg⁻¹), and explain the comparisons. (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg, mₙ = 1.675 × 10−27 kg, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – (a) count the particles
2 protons, 2 neutrons: Q = 2 × 1.60 × 10−19 = 3.20 × 10−19 C; m = 2 × 1.673 × 10−27 + 2 × 1.675 × 10−27 = 6.696 × 10−27 kg
F: specific charge = Q ÷ m
I: specific charge = 3.20 × 10−19 ÷ 6.696 × 10−27
A: 4.778… × 107 = 4.78 × 107 C kg⁻¹ (3 s.f.)
Step 2 – (b) comparison
the alpha particle and ¹²C⁶⁺ have the same specific charge: each has equal numbers of protons and neutrons, so both charge and mass are 3 times bigger for carbon and the ratio is unchanged
proton ÷ alpha = 9.563… × 107 ÷ 4.778… × 107 = 2.00 (3 s.f.): the alpha particle has twice the charge but about four times the mass, so its specific charge is about half the proton’s
Q7 (Exam) Calculate the specific charge of a uranium-235 nucleus, ²³⁵₉₂U, to 3 significant figures. Explain why it is smaller than that of a ¹²₆C nucleus (4.78 × 107 C kg⁻¹). (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg, mₙ = 1.675 × 10−27 kg, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – count the particles
92 protons and 235 − 92 = 143 neutrons
Q = 92 × 1.60 × 10−19 = 1.472 × 10−17 C
m = 92 × 1.673 × 10−27 + 143 × 1.675 × 10−27 = 1.53916 × 10−25 + 2.39525 × 10−25 = 3.93441 × 10−25 kg
Step 2 – specific charge
F: specific charge = Q ÷ m
I: specific charge = 1.472 × 10−17 ÷ 3.93441 × 10−25
F: specific charge is already the subject
A: 3.741… × 107 = 3.74 × 107 C kg⁻¹ (3 s.f.)
heavy nuclei have more neutrons than protons (143 neutrons for 92 protons), so they have more mass per unit charge than ¹²C, which has equal numbers
Watch out: using the nucleon number 235 as the number of neutrons. Neutrons = A − Z = 143.
Q8 (Exam) An oxygen atom, ¹⁶₈O, gains two electrons to form an O²⁻ ion. Calculate the magnitude of the specific charge of the ion, to 3 significant figures, and state the sign of its charge. Show that including the electrons’ mass makes no difference at this precision. (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg, mₙ = 1.675 × 10−27 kg, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – count the particles
8 protons, 8 neutrons and 8 + 2 = 10 electrons; charge = (8 − 10) × e = −2e, magnitude 3.20 × 10−19 C
nucleus: 8 × 1.673 × 10−27 + 8 × 1.675 × 10−27 = 2.6784 × 10−26 kg; electrons: 10 × 9.11 × 10−31 = 9.11 × 10−30 kg
m = 2.6784 × 10−26 + 0.000911 × 10−26 = 2.6793… × 10−26 kg
Step 2 – specific charge
F: specific charge = Q ÷ m
I: specific charge = 3.20 × 10−19 ÷ 2.6793… × 10−26
F: specific charge is already the subject
A: 1.1943… × 107 = 1.19 × 107 C kg⁻¹ (3 s.f.); the charge is negative
ignoring the electrons: 3.20 × 10−19 ÷ 2.6784 × 10−26 = 1.1947… × 107 C kg⁻¹, also 1.19 × 107 C kg⁻¹, so the electrons’ mass is negligible
Watch out: using a charge of 8e or 10e. The ion’s charge depends only on the excess electrons: 10 electrons − 8 protons = 2 extra, so −2e.
Q9 (Stretch) A nucleus contains 8 neutrons and has a specific charge of 4.46 × 107 C kg⁻¹. Calculate its proton number and identify the nuclide. (e = 1.60 × 10−19 C, mₚ = 1.673 × 10−27 kg, mₙ = 1.675 × 10−27 kg, mₑ = 9.11 × 10−31 kg)
Show solution
Step 1 – write the specific charge in terms of Z
F: specific charge = Q ÷ m = Ze ÷ (Zmₚ + 8mₙ)
I: 4.46 × 107 = Z × 1.60 × 10−19 ÷ (Z × 1.673 × 10−27 + 8 × 1.675 × 10−27)
Step 2 – rearrange for Z
F: 4.46 × 107 × (Z × 1.673 × 10−27 + 1.340 × 10−26) = Z × 1.60 × 10−19, so Z = (4.46 × 107 × 1.340 × 10−26) ÷ (1.60 × 10−19 − 4.46 × 107 × 1.673 × 10−27)
Z = 5.9764 × 10−19 ÷ (1.60 × 10−19 − 7.4615… × 10−20) = 5.9764 × 10−19 ÷ 8.5384… × 10−20 = 6.999…
A: Z = 7, so A = 7 + 8 = 15: the nuclide is nitrogen-15, ¹⁵₇N
Check: 7 × 1.60 × 10−19 ÷ (7 × 1.673 × 10−27 + 8 × 1.675 × 10−27) = 1.12 × 10−18 ÷ 2.5111 × 10−26 = 4.46 × 107 C kg⁻¹ (3 s.f.) ✓
Q10 (Stretch) In a mass spectrometer, a copper-63 ion of mass 62.9 u is found to have a specific charge of 3.06 × 106 C kg⁻¹. (a) Calculate the charge on the ion and deduce how many electrons were removed. (b) Copper has proton number 29. State the number of protons, neutrons and electrons in the ion. (1 u = 1.661 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: m = 62.9 × 1.661 × 10−27 = 1.0447… × 10−25 kg
Step 1 – (a) charge
F: specific charge = Q ÷ m
I: 3.06 × 106 = Q ÷ 1.0447… × 10−25
F: Q = 3.06 × 106 × 1.0447… × 10−25 = 3.196… × 10−19 C
number of electrons removed = 3.196… × 10−19 ÷ 1.60 × 10−19 = 1.99… ≈ 2
A: Q = 3.20 × 10−19 C (3 s.f.), so 2 electrons were removed: the ion is Cu²⁺
Step 2 – (b) particles
A: 29 protons, 63 − 29 = 34 neutrons, 29 − 2 = 27 electrons
Watch out: forgetting to convert u to kg. Using m = 62.9 gives a charge of about 2 × 108 C, which is clearly impossible for one ion.
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