Discovery of electron (Thomson, cathode rays), proton (Goldstein), neutron (Chadwick): Worked examples

This page drills the numbers behind the parent topic. The parent gave you the ideas and the master formula. Here we grind through every kind of number the topic can hand you — every sign, every degenerate case, every limit, a word problem, and an exam trap.

Before we start, one promise: no symbol is used before it is named. Let us list the players once.

Definition The symbols we will use (all in SI)

$e$ = size of one elementary charge $= 1.602 \times 1 0^{- 19}$ coulomb (C). A coulomb is just "a bucket of charge."
$m$ = mass of the particle, in kilograms (kg).
$\frac{e}{m}$ = charge-to-mass ratio — "how much charge you get per kilogram." Big ratio = lots of charge on very little mass = a light charged thing. See Charge to Mass Ratio.
$R$ = a shorthand name we will give to a measured value of $e / m$ (so we can do algebra with it). Same units, C/kg.
$E$ = electric field strength (volts per metre, V/m) — "how hard the plates push a charge."
$B$ = magnetic field strength (tesla, T) — "how hard a magnet twists a moving charge."
$v$ = speed of the beam (metres per second, m/s).
$L$ = length of the deflecting plates (m); $y$ = how far the beam bends sideways (m).
$F$ = force (newtons, N); $a$ = sideways acceleration (m/s², "how fast the sideways speed grows"); $t$ = transit time, the seconds the particle spends between the plates.

The master formula, built once (so nothing is a black box)

Every $e / m$ example below leans on one equation. Let us earn it now with plain mechanics, so it is never a magic quote.

Build-Step 1 — sideways force. The field $E$ pushes the charge $e$ with force $F = e E$ . Why? That is the definition of an electric field: force per unit charge, so force $=$ charge $\times$ field.

Build-Step 2 — turn force into acceleration. Newton's law $F = ma$ gives $a = \frac{F}{m} = \frac{e E}{m} .$ Why? We want motion, and acceleration is what a force produces on a mass $m$ .

Build-Step 3 — how long is the push on? The particle travels the plate length $L$ at forward speed $v$ , so the transit time is $t = \frac{L}{v} .$ Why? Time $=$ distance $\div$ speed. The sideways push acts only while the particle is between the plates.

Build-Step 4 — sideways distance fallen. For constant acceleration from rest sideways, $y = \frac{1}{2} a t^{2} = \frac{1}{2} \cdot \frac{e E}{m} \cdot \frac{L ^{2}}{v ^{2}} .$ Why? This is the standard "distance $= \frac{1}{2} a t^{2}$ " for something starting with zero sideways speed.

Build-Step 5 — solve for the ratio. Rearrange to isolate $e / m$ : $\frac{e}{m} = \frac{2 y v ^{2}}{E L ^{2}}$ Why? Everything on the right is measurable; the ratio $e / m$ is the one unknown. This is the workhorse for Ex 1 and Ex 7.

The scenario matrix

Every problem in this topic is one of these cells. Each worked example below is tagged with the cell it kills.

#	Case class	What is unusual about it	Killed by
A	Standard $e / m$ from balanced fields	The textbook 3-step Thomson chain	Ex 1
B	Get mass from $e / m$ + Millikan $e$	Two experiments combined	Ex 2
C	Sign of the charge (positive vs negative)	Which way does the beam bend?	Ex 3
D	Gas-dependent $e / m$ (canal rays)	Answer changes with the gas — not universal	Ex 4
E	Degenerate / limiting input ( $B \to 0$ , $E \to 0$ , or $E = v B$ balance)	Formula blows up, or straight line, or circular	Ex 5
F	Neutron by missing mass (nuclear bookkeeping)	Charge already balances, mass does not	Ex 6
G	Real-world word problem	Numbers hidden in a story (TV tube)	Ex 7
H	Exam twist / trap	Ratio comparison, no calculator needed	Ex 8

Ex 1 — Cell A: the full Thomson chain

Step 1 — Find the speed $v$ from the balance. $v = \frac{E}{B} = \frac{2.0 \times 1 0 ^{4}}{1.0 \times 1 0 ^{- 3}} = 2.0 \times 1 0^{7} m/s$ Why this step? When the beam goes straight, the electric push $e E$ exactly cancels the magnetic push $e v B$ . The $e$ cancels on both sides, so $v = E / B$ has no mass or charge in it — it is pure, measured field strengths. That is why Thomson did the balance first: it isolates $v$ .

Step 2 — Use the master formula built above: $\frac{e}{m} = \frac{2 y v ^{2}}{E L ^{2}}$ Why this step? We derived this from $F = e E$ , $a = e E / m$ , $t = L / v$ , $y = \frac{1}{2} a t^{2}$ . Everything on the right is now a known number, so we can solve for the ratio.

Step 3 — Plug in carefully. Note $L^{2} = (0.05)^{2} = 2.5 \times 1 0^{- 3} m^{2}$ . $\frac{e}{m} = \frac{2 ( 2.5 \times 1 0 ^{- 3} ) ( 2.0 \times 1 0 ^{7} ) ^{2}}{( 2.0 \times 1 0 ^{4} ) ( 2.5 \times 1 0 ^{- 3} )} = \frac{2.0 \times 1 0 ^{12}}{50} = 4.0 \times 1 0^{10} C/kg$

Verify: Units — $\frac{m \cdot ( m/s ) ^{2}}{( V/m ) \cdot m ^{2}}$ . Since $1 V/m = 1 N/C$ , the whole thing collapses to C/kg. ✓ The answer $4.0 \times 1 0^{10}$ is the same order of magnitude as the real $1.758 \times 1 0^{11}$ (these are made-up lab numbers, not a real electron run) — matches the forecast of "huge." ✓

Ex 2 — Cell B: mass from two experiments

Step 1 — See what each experiment owns. Thomson owns the ratio $e / m$ ; Millikan owns the charge $e$ . Mass is $m$ . Let us name the measured ratio $R = e / m = 1.758 \times 1 0^{11} C/kg$ (introduced in the symbol table) so the algebra is clean.

Step 2 — Algebra. From $\frac{e}{m} = R$ we get $m = \frac{e}{R}$ . Why this step? $R$ has $e$ on top and $m$ on the bottom. Dividing the stand-alone $e$ by $R$ makes the two $e$ 's cancel, leaving $1/ m$ flipped up to $m$ . This is the entire reason both experiments were needed.

Step 3 — Plug in. $m = \frac{e}{R} = \frac{1.602 \times 1 0 ^{- 19}}{1.758 \times 1 0 ^{11}} = 9.11 \times 1 0^{- 31} kg$

Verify: Units: $\frac{C}{C/kg} = kg$ . ✓ Matches the accepted electron mass $9.11 \times 1 0^{- 31}$ kg. ✓

Ex 3 — Cell C: which way does it bend?

Figure — Discovery of electron (Thomson, cathode rays), proton (Goldstein), neutron (Chadwick)

Figure caption (read this first): The figure shows the beam entering from the left along the dashed centre line. The yellow horizontal bars are the plates — the top one is marked $+$ , the bottom one $-$ . The blue curve is a negative particle: it arcs upward toward the $+$ plate (blue arrow). The pink curve is a positive particle: it arcs downward toward the $-$ plate (pink arrow). Colours are only labels for the two signs; the shapes alone tell the story.

Step 1 — State the rule. A positive plate attracts negative charge and repels positive charge. In the figure the blue beam curves up toward the $+$ , the pink beam curves down toward the $-$ .

Step 2 — Read the case. Bends up toward the + plate ⇒ it is pulled to positive ⇒ the particle is negative (an electron / cathode ray). Why this step? This is literally how Thomson knew cathode rays were negative — no formula, just the direction of bend.

Step 3 — The mirror case. Bends down toward the – plate ⇒ pulled to negative ⇒ the particle is positive (a canal ray / ion). This is Goldstein's rays going the opposite way.

Verify: Sanity check — cathode rays (–) and canal rays (+) must bend in opposite directions in the same field. The figure shows exactly that. ✓

Ex 4 — Cell D: the gas-dependent ratio

Step 1 — Write the ratio for each ion. Both carry charge $+ e$ . Proton mass $m_{p}$ ; helium ion mass $\approx 4 m_{p}$ . $(\frac{e}{m})_{H} = \frac{e}{m _{p}}, (\frac{e}{m})_{He} = \frac{e}{4 m _{p}}$

Step 2 — Take the ratio of ratios. $\frac{( e / m ) _{He}}{( e / m ) _{H}} = \frac{e / ( 4 m _{p} )}{e / m _{p}} = \frac{1}{4}$ Why this step? The charge $e$ cancels, so the ratio depends only on mass — this is precisely why canal-ray $e / m$ is not universal (see Isotopes and Mass Number: heavier nuclei, smaller $e / m$ ).

Step 3 — Compute. $(\frac{e}{m})_{He} = \frac{9.58 \times 1 0 ^{7}}{4} = 2.395 \times 1 0^{7} C/kg$

Verify: Heavier ion ⇒ smaller ratio. ✓ Exactly $\frac{1}{4}$ of hydrogen's, matching the mass factor 4. ✓

[!mistake] Do not reuse the electron's $1.758 \times 1 0^{11}$ here. That is universal for electrons only. Positive rays are ionised gas, so each gas gives its own number.

Ex 5 — Cell E: degenerate / limiting inputs

Step 1 — Case $B \to 0$ (electric only). The magnetic push $e v B$ shrinks to zero, so only the electric push $F = e E$ remains, giving sideways acceleration $a = e E / m$ . There is nothing left to cancel it. Why this step? The balance condition $v = E / B$ has $B$ in the denominator. As $B \to 0$ , $E / B \to \infty$ — meaning no finite speed can be balanced; the beam simply bends fully under $E$ with the parabola $y = \frac{1}{2} a t^{2}$ . The formula $v = E / B$ does not apply here (division by zero); this is the deflect-with- $E$ -only stage.

Step 2 — Case perfect balance $e E = e v B$ . Cancel $e$ (allowed, $e \neq = 0$ ), then divide out $v$ : $E = v B \Rightarrow v = \frac{E}{B}$ Net sideways force $= e E - e v B = e (E - v B) = 0$ , so $a = 0$ and the transit-time projectile formula gives $y = \frac{1}{2} a t^{2} = 0$ . Why this step? Zero net force ⇒ zero sideways acceleration ⇒ zero deflection. The beam goes dead straight — this is exactly the setup Thomson uses to read off $v$ .

Step 3 — Case $E \to 0$ (magnetic only). Now only the magnetic force $F = e v B$ acts. A magnetic force is always perpendicular to the velocity, so it never speeds the particle up — it only turns it. Constant-size force always at right angles to motion ⇒ the path is a circle of radius $r = \frac{m v}{e B}$ . Why this step? This is the third degenerate member of the same family: no electric push, so no parabola and no straight line — a pure bend. It is the principle behind the mass spectrometer.

Verify: Plug $E = v B$ into net force: $e (E - v B) = e \cdot 0 = 0$ , so $y = 0$ . ✓ Radius formula units: $\frac{kg \cdot m/s}{C \cdot T} = m$ (since $1 T = 1 kg / (C \cdot s)$ ). ✓ Three opposite extremes all handled.

Ex 6 — Cell F: the neutron by missing mass

Step 1 — Account for charge. 2 protons ( $+ 2$ ) $+$ 2 electrons ( $- 2$ ) $= 0$ . Charge is fully explained — nothing charged is missing. Why this step? This is the whole logic of Chadwick's neutron: the charge books already balance, so any leftover mass must be uncharged.

Step 2 — Account for mass. $missing mass = 4 u - 2 u (protons) - (electrons \approx 0) = 2 u$

Step 3 — Convert to particles. Each neutron $\approx 1$ u, so $number of neutrons = \frac{2 u}{1 u} = 2.$ So helium is 2 protons + 2 neutrons — see Rutherford Nuclear Model for where they sit.

Verify: 2 p $+$ 2 n $= 2 + 2 = 4$ u total. ✓ Charge $= + 2 - 2 = 0$ . ✓ Both books balance simultaneously — that is what a real particle must satisfy.

Ex 7 — Cell G: real-world word problem (old TV tube)

Step 1 — Rearrange the master formula for $y$ . From $\frac{e}{m} = \frac{2 y v ^{2}}{E L ^{2}}$ : $y = \frac{( e / m ) E L ^{2}}{2 v ^{2}}$ Why this step? Here we know the ratio and want the geometry $y$ — the reverse of Ex 1. Same equation we built at the top, different unknown.

Step 2 — Plug in numbers. $y = \frac{( 1.758 \times 1 0 ^{11} ) ( 5.0 \times 1 0 ^{4} ) ( 0.04 ) ^{2}}{2 ( 3.0 \times 1 0 ^{7} ) ^{2}}$

Step 3 — Compute. Numerator $= 1.758 \times 1 0^{11} \times 5.0 \times 1 0^{4} \times 1.6 \times 1 0^{- 3} = 1.4064 \times 1 0^{13}$ . Denominator $= 2 \times 9.0 \times 1 0^{14} = 1.8 \times 1 0^{15}$ . $y = \frac{1.4064 \times 1 0 ^{13}}{1.8 \times 1 0 ^{15}} = 7.81 \times 1 0^{- 3} m \approx 7.8 mm$

Verify: A few millimetres of bend — exactly the scale a TV needs to sweep the dot across a screen. ✓ Units: $\frac{( C/kg ) ( V/m ) ( m ^{2} )}{( m/s ) ^{2}} = m$ . ✓

Ex 8 — Cell H: exam twist (no calculator)

Step 1 — Same charge, so divide the ratios. $\frac{( e / m ) _{e}}{( e / m ) _{p}} = \frac{e / m _{e}}{e / m _{p}} = \frac{m _{p}}{m _{e}}$ Why this step? Equal $e$ cancels, leaving the inverse of the mass ratio — the proton-to-electron mass ratio, a famous number.

Step 2 — Estimate the division. $\frac{1.758 \times 1 0 ^{11}}{9.58 \times 1 0 ^{7}} \approx \frac{1.76}{9.58} \times 1 0^{4} \approx 0.1836 \times 1 0^{4} = 1836$

Step 3 — Interpret. The factor is $\approx 1836$ = the proton is 1836 times heavier than the electron. The parent note's " $1836 m_{e}$ " falls straight out.

Verify: $9.58 \times 1 0^{7} \times 1836 \approx 1.758 \times 1 0^{11}$ . ✓ The mystery factor is a mass ratio, not a charge ratio — the classic trap. ✓

Recall

When do you divide $e$ by $e / m$ ? ::: To get mass $m$ (Ex 2, Cell B). Canal-ray $e / m$ for He $^{+}$ vs H $^{+}$ ? ::: One quarter, because He $^{+}$ is $4 \times$ heavier (Ex 4, Cell D). Perfect balance $E = v B$ gives what deflection? ::: Zero — a straight beam (Ex 5, Cell E). With $E \to 0$ and only $B$ on, what is the path? ::: A circle of radius $r = m v / (e B)$ (Ex 5, Cell E). Helium's missing mass corresponds to what? ::: 2 neutrons (Ex 6, Cell F). Ratio of electron to proton $e / m$ equals what physical number? ::: $m_{p} / m_{e} \approx 1836$ (Ex 8, Cell H).

Connections

Charge to Mass Ratio — every example above is an application of this one quantity.
Millikan Oil Drop Experiment — supplies the $e$ used in Ex 2.
Isotopes and Mass Number — why canal-ray $e / m$ (Ex 4) and neutron counting (Ex 6) matter.
Rutherford Nuclear Model — where the protons and neutrons of Ex 6 actually sit.
Thomson Plum Pudding Model — the model built on the electron of Ex 1–3.