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

Straight from the summary table.

• (a) Cathode rays → electron, discovered by Thomson, charge negative ($-1.602\times 10^{-19}$ C).
• (b) Canal rays → proton, discovered by Goldstein, charge positive ($+1.602\times 10^{-19}$ C).
• (c) Be + $\alpha$ → neutron, discovered by Chadwick, charge zero. Mnemonic: Elephants Push Nuts (Electron→Proton→Neutron).

They bend toward the positive plate. Opposite charges attract, so the rays must carry the opposite sign to the positive plate — they are negative. This is the direct evidence that cathode rays are negatively charged.

He measured the charge-to-mass ratio of the electron, not the charge or mass separately. $\frac{e}{m}=1.758\times 10^{11}{\text{C kg}}^{-1}$

WHAT: "Passes straight through" means the electric push and the magnetic push are equal and opposite — they cancel. WHY this matters: when they balance, the charge $e$ and mass $m$ drop out entirely, leaving only the fields and the speed. $eE=evB\Rightarrow v=\frac{E}{B}$ $v=\frac{3.0\times 10^4}{2.0\times 10^{-3}}=1.5\times 10^7{\text{m s}}^{-1}$ What it looks like (figure): the blue electric-force arrow and the orange magnetic-force arrow are the same length and point opposite ways, so the beam (green) stays flat.

WHY divide? The ratio tells us $e/m$; if we already know $e$, then $m=e÷(e/m)$ recovers the mass alone. $m=\frac{e}{e/m}=\frac{1.602\times 10^{-19}}{1.758\times 10^{11}}=9.11\times 10^{-31}\text{kg}$ This is the famous electron mass.

$\frac{e}{m}{|}_{\text{proton}}=\frac{1.602\times 10^{-19}}{1.672\times 10^{-27}}=9.58\times 10^7{\text{C kg}}^{-1}$ Compare with the electron: $\frac{1.758\times 10^{11}}{9.58\times 10^7}\approx 1836$ The proton's $e/m$ is about 1836 times smaller, because the proton is about 1836 times heavier (same size of charge).

The positive canal-ray particle is an ionised gas atom, so its mass equals the atom's mass. All these ions carry (roughly) the same size of charge, so: $\frac{e}{m}\propto \frac{1}{m}.$ Smaller mass ⟺ larger ratio. Hydrogen gives the smallest ion (a bare proton), so it gives the largest $e/m$. Switch to helium, mass ≈ 4× larger ⟹ the ratio drops to roughly one quarter of the hydrogen value.

• Cathode rays: the particle (electron) comes from the tube's electric current itself, and electrons are identical everywhere — so one universal $e/m$. This universality is the whole reason we say the electron is a fundamental building block (see Thomson Plum Pudding Model).
• Canal rays: the particle is made by ionising the gas atoms, so its mass = that gas's atomic mass, which obviously differs from gas to gas.

WHAT: Charge already balances at 2 units, so we cannot add more charged particles without breaking the charge count. WHY a new particle: the extra ~2 mass-units must come from something with mass but no charge — the neutron (Chadwick). Adding protons would ruin the charge; adding electrons adds negligible mass; only a neutral, proton-heavy particle fits. See Isotopes and Mass Number for how varying neutron count creates isotopes.

Step 1 — speed from the balanced beam. WHAT: use the undeflected condition; WHY: it isolates $v$ from just the fields. $v=\frac{E}{B}=\frac{4.0\times 10^4}{1.0\times 10^{-3}}=4.0\times 10^7{\text{m s}}^{-1}$ Step 2 — turn the deflection formula around. The beam is now projectile motion: steady forward speed $v$, steady sideways push. From $y=\frac{1}{2}\frac{eE}{m}\frac{L^2}{v^2}$ solve for the ratio: $\frac{e}{m}=\frac{2y{v}^2}{E{L}^2}$ Step 3 — substitute. $\frac{e}{m}=\frac{2(5.0\times 10^{-3})(4.0\times 10^7{)}^2}{(4.0\times 10^4)(5.0\times 10^{-2}{)}^2}=1.6\times 10^{11}{\text{C kg}}^{-1}$ Close to Thomson's $1.758\times 10^{11}$ — the figure shows the parabolic bend once $B$ is off (green curve).

$m=\frac{e}{e/m}=\frac{1.602\times 10^{-19}}{1.6\times 10^{11}}=1.00\times 10^{-30}\text{kg}$ That is about $1.0\times 10^{-30}$ kg, a bit heavier than the true $9.11\times 10^{-31}$ kg. Why the difference: our $e/m$ came out slightly lower than Thomson's real value (rounded fields/deflection), and since $m=e÷(e/m)$, a smaller ratio gives a larger mass. Experimental rounding, not new physics.

True. The selector passes only the speed where forces balance: $v=E/B$. With $E$ fixed and $B$ doubled, $v_{\text{new}}=\frac{E}{2B}=\frac{1}{2}\frac{E}{B}=\frac{1}{2}{v}_{\text{old}}.$ Edge case — faster particles: any particle moving faster than $v_{\text{new}}$ now feels a magnetic force $evB$ that is bigger than the electric force $eE$, so it gets deflected out of the beam. The selector always keeps only the one speed $E/B$; everything else curves away. This is exactly why it is called a selector.

• Cathode-ray $e/m$: unchanged. The particle is the electron, which is identical regardless of gas — universal $e/m=1.758\times 10^{11}{\text{C kg}}^{-1}$.
• Canal-ray $e/m$: changes (drops). The positive particle is now Ne⁺, whose mass ≈ 20 proton-masses versus hydrogen's 1. Since $e/m\propto 1/m$, the ratio falls to roughly $1/20$ of the hydrogen value. One-line rule: electrons come from the electricity; positive ions come from the gas.

Balance the top numbers (mass number A): $9+4=A+1\Rightarrow A=12$. Balance the bottom numbers (charge/atomic number Z): $4+2=Z+0\Rightarrow Z=6$. So the carbon is ${}_6^{12}\text{C}$, and the emitted particle is ${}_0^{1}n$ — mass number 1, charge 0. Why it must be neutral: the charge column already balances ($6=4+2$), so the extra particle carries zero charge — consistent with the fact that the real radiation was undeflected by electric and magnetic fields. Why ~proton-mass: the mass column needs 1 unit for it, matching Chadwick's momentum-conservation finding that it knocked protons out of paraffin with comparable mass. See Rutherford Nuclear Model for where this neutron then lives.

• (a) $E\to 0$: $y\to 0$. No electric field, no sideways push, beam stays straight. ✔ makes sense.
• (b) $v\to \infty$: $y\to 0$ (it sits in the denominator as $v^2$). A blisteringly fast particle spends almost no time between the plates ($t=L/v\to 0$), so the tiny sideways push has no time to act. ✔ makes sense. Both limits confirm the formula behaves sanely at its edges — a good sanity check any time you derive a projectile-style deflection.

• I can state each particle, discoverer, charge, and mass from memory ::: L1
• I can compute $v=E/B$ and $e/m=2y{v}^2/(E{L}^2)$ with correct units ::: L2, L4
• I can explain why cathode-ray $e/m$ is universal but canal-ray $e/m$ isn't ::: L3
• I can balance the Chadwick reaction and justify a neutral, proton-mass particle ::: L5
• I know the velocity selector filters by speed, not by charge or mass ::: L5