Determining Specific Charge Revision Notes | A Level AQA Physics

1. The Mystery of Cathode Rays

In the 1890s, physicists knew that sealed glass tubes with metal electrodes at each end produced a glowing beam when a high voltage was applied. The beam travelled from the negative electrode (cathode) to the positive electrode (anode) — but nobody knew what the beam actually was. Were these rays a form of electromagnetic radiation, or were they particles? And if particles, what kind?

J.J. Thomson set out to answer this in 1897. His approach was elegant: if the beam consisted of charged particles, electric and magnetic fields should deflect it in predictable ways. By measuring how much the beam deflected under known fields, he could calculate a fundamental property of the particles — their specific charge, $e/m_e$.

2. Specific Charge — What It Is and Why It Matters

Notice that specific charge is a ratio — you do not need to know $e$ and $m_e$ individually to measure it. Thomson could determine $e/m_e$ purely from the deflection of the beam, without ever isolating a single electron. That is what made his experiment so powerful.

3. Thomson's Method — The Crossed-Fields Experiment

Picture a cathode ray tube with two additional features: a pair of parallel plates that create a uniform electric field $E$ perpendicular to the beam, and a pair of coils that create a uniform magnetic field $B$ also perpendicular to the beam (but at right angles to $E$). Thomson used these two fields together in a two-stage strategy.

In the first stage, both fields are applied simultaneously and adjusted until the beam travels straight through — no deflection. At this point the electric force and the magnetic force on each particle are equal and opposite. This is the velocity selector condition.

In the second stage, the electric field is switched off, leaving only the magnetic field. The beam now curves in a circular arc because the magnetic force acts as a centripetal force. By measuring the radius of curvature $r$, Thomson could write down a second equation.

Substituting $v = E/B$ into $e/m_e = v/(Br)$ gives the fully combined result: $e/m_e = E/(B^2 r)$. Every quantity on the right-hand side — $E$, $B$, and $r$ — is directly measurable in the laboratory. That is the elegance of Thomson's approach.

4. Worked Calculation — Finding Specific Charge

Theory is only useful when you can apply it to numbers. The following worked example walks through a standard AQA-style calculation, showing every substitution and unit check explicitly.

Worked example: Calculating the specific charge of the electron from Thomson's experiment

In a Thomson-type experiment, the electric field strength is $E = 3.0 \times 10^4$ V m$^{-1}$ and the magnetic flux density is $B = 1.5 \times 10^{-3}$ T. When only the magnetic field acts, the beam follows a circular arc of radius $r = 0.12$ m. Calculate the specific charge $e/m_e$.

1. $v = \frac{E}{B} = \frac{3.0 \times 10^4}{1.5 \times 10^{-3}}$ — When the beam is undeflected, $eE = evB$, so $v = E/B$. Both $E$ and $B$ are given directly. [M1]
2. $v = \frac{3.0 \times 10^4}{1.5 \times 10^{-3}} = 2.0 \times 10^7 \text{ m s}^{-1}$ — Dividing: $3.0 \div 1.5 = 2.0$ and $10^4 \div 10^{-3} = 10^{4-(-3)} = 10^7$. Speed is $2.0 \times 10^7$ m s$^{-1}$. [A1]
3. $\frac{e}{m_e} = \frac{v}{Br}$ — From $evB = m_e v^2 / r$, dividing both sides by $m_e v$ gives this result. [M1]
4. $\frac{e}{m_e} = \frac{2.0 \times 10^7}{(1.5 \times 10^{-3}) \times 0.12}$ — Substituting $v = 2.0 \times 10^7$ m s$^{-1}$, $B = 1.5 \times 10^{-3}$ T, $r = 0.12$ m. Carry units through.
5. $(1.5 \times 10^{-3}) \times 0.12 = 1.5 \times 0.12 \times 10^{-3} = 0.18 \times 10^{-3} = 1.8 \times 10^{-4}$ — $1.5 \times 0.12$: multiply $15 \times 12 = 180$, so $1.5 \times 0.12 = 0.18$. Then $0.18 \times 10^{-3} = 1.8 \times 10^{-4}$.
6. $\frac{e}{m_e} = \frac{2.0 \times 10^7}{1.8 \times 10^{-4}} = \frac{2.0}{1.8} \times 10^{7-(-4)} = 1.11 \times 10^{11} \text{ C kg}^{-1}$ — $2.0 \div 1.8 \approx 1.11$ and $10^7 \div 10^{-4} = 10^{11}$. The unit is C kg$^{-1}$. [A1] This is close to the accepted value of $1.76 \times 10^{11}$ C kg$^{-1}$ — the small discrepancy reflects the simplified numbers used here.

Answer: $e/m_e \approx 1.11 \times 10^{11}$ C kg$^{-1}$

5. The Significance of Thomson's Result

When Thomson obtained his value of $e/m_e$, the result was startling. The specific charge was roughly 1800 times larger than the specific charge of the hydrogen ion, which was already known from electrolysis experiments. Since the charges were expected to be of similar magnitude, this implied that the mass of the cathode ray particle was far smaller than the mass of the lightest known atom — hydrogen.

This was the first direct experimental evidence that atoms are not the smallest constituents of matter. The cathode ray particles — which Thomson called corpuscles, later renamed electrons — were sub-atomic. Thomson's measurement therefore marked a genuine turning point in physics: the discovery of the electron as a particle smaller than any atom.

A further implication was that the same $e/m_e$ value was obtained regardless of the cathode material or the residual gas in the tube. This universality showed that electrons are a common constituent of all matter, not an artefact of one particular substance.

6. Comparing the Specific Charge of the Electron and the Hydrogen Ion

The hydrogen ion $\text{H}^+$ is a proton — a hydrogen atom that has lost its single electron. Its charge is $+e = 1.60 \times 10^{-19}$ C and its mass is $m_p = 1.67 \times 10^{-27}$ kg. The specific charge of the hydrogen ion is therefore $e/m_p = 1.60 \times 10^{-19} / 1.67 \times 10^{-27} \approx 9.58 \times 10^7$ C kg$^{-1}$.

Both particles carry the same magnitude of charge. The ratio of their specific charges is therefore equal to the inverse ratio of their masses: $m_p / m_e \approx 1836$. The electron's specific charge is about 1836 times larger than that of the hydrogen ion, confirming that the electron is roughly 1836 times less massive than the proton.

This comparison is what made Thomson's result so revolutionary. Before 1897, the hydrogen ion had the largest known specific charge of any particle. Thomson's cathode ray particles shattered that record by a factor of nearly 2000, pointing unambiguously to a new, far lighter particle.

7. The Practical — Measuring Specific Charge in the Laboratory

Notice that the fine-beam tube method uses energy conservation ($eV = \frac{1}{2}m_e v^2$) to find the electron speed, rather than a velocity selector. This is an alternative but equally valid route to $e/m_e$, and AQA requires familiarity with at least one method.

8. Exam Technique and Common Mistakes

AQA questions on this topic tend to fall into three categories: calculation questions requiring you to apply $v = E/B$ and $e/m_e = v/(Br)$ or $e/m_e = 2V/(B^2r^2)$; explain/describe questions asking about the significance of Thomson's result; and compare questions asking you to contrast the electron and hydrogen ion specific charges. Each type rewards a slightly different approach.