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Inquiry Question 2: How is it known that atoms are made up of protons, neutrons and electrons?

Investigate, assess and model the experimental evidence supporting the existence and properties of the electron, including cathode ray tube experiments and Thomson's determination of the charge-to-mass ratio of the electron

A focused answer to the HSC Physics Module 8 dot point on the discovery and properties of the electron. Cathode ray tubes and the particle vs wave debate, Thomson's crossed-field experiment to measure the charge-to-mass ratio e/m, and his plum-pudding model of the atom.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context

What this dot point is asking

NESA wants you to describe the cathode ray tube experiments of the late 19th century, summarise the evidence that cathode rays are negatively charged particles (later called electrons), and describe Thomson's apparatus and reasoning by which he measured the charge-to-mass ratio e/me/m. You should also state Thomson's plum-pudding model of the atom as the model that the discovery of the electron immediately suggested.

The answer

Cathode ray tubes

Cathode ray tube with crossed electric and magnetic fields A sealed tube with a cathode and anode on the left producing a beam of electrons. The beam passes between two horizontal deflection plates that create a uniform electric field pointing downward, with a magnetic field region marked by crosses indicating a field into the page. With both fields balanced the beam travels straight to the centre of the screen on the right. With the magnetic field switched off the electric field alone bends the beam upward to a deflected spot near the top of the screen. cathode anode + plate (upper) - plate (lower) E B into page (field region between the plates) undeflected (fields balanced): v = E ⁄ B B off: E alone deflects centre spot deflected spot, y high V

A cathode ray tube is a sealed glass tube containing two electrodes and a low-pressure gas. A high voltage applied between the cathode (negative electrode) and anode (positive electrode) produces a stream of "cathode rays" travelling from cathode to anode. The rays make the residual gas glow and produce fluorescence on a screen at the far end of the tube.

By the 1890s the question was: what are cathode rays? Two camps:

  • the wave camp (mainly German physicists) thought cathode rays were a wave phenomenon in the aether, somewhat like light,
  • the particle camp (mainly British physicists) thought they were streams of charged particles.

Evidence for particles

Several observations pointed to particles:

  • A small obstacle placed in the beam casts a sharp shadow, consistent with straight-line travel.
  • A small paddle wheel placed in the path is set spinning, indicating that the rays carry momentum.
  • A magnetic field deflects the rays along a curved path, with direction consistent with negatively charged particles.
  • An electric field (between two parallel plates) deflects them in the direction expected of negative charges.
  • They are emitted in any direction from the cathode (regardless of where the anode is), suggesting emission from the cathode metal itself.

The deflection by an electric field was particularly damning for the wave model: an electromagnetic wave carries no net charge and is not deflected by static E\vec{E}.

Thomson's crossed-field experiment (1897)

J. J. Thomson designed an apparatus to measure properties of the cathode rays themselves. Inside the tube he added two horizontal parallel plates creating a uniform vertical electric field E\vec{E}, and a pair of coils producing a horizontal magnetic field B\vec{B} perpendicular to both E\vec{E} and to the beam direction. The combination is a "velocity selector":

  • Electric force on an electron moving along the beam: FE=qEF_E = qE (vertical).
  • Magnetic force: FB=qvBF_B = qvB (vertical, opposite to FEF_E when E\vec{E} and B\vec{B} are arranged correctly).

When the two forces are balanced, the beam passes undeflected. Setting qE=qvBqE = qvB gives:

v=EBv = \frac{E}{B}

This is one equation in the wanted ratios, independent of qq and mm.

Thomson then switched off B\vec{B} and measured the vertical deflection yy caused by E\vec{E} alone over the plate length LL. The electron experiences acceleration a=qE/ma = qE/m for a time t=L/vt = L/v while between the plates. The deflection is:

y=12at2=12qEmL2v2y = \tfrac{1}{2} a t^2 = \tfrac{1}{2} \frac{q E}{m} \frac{L^2}{v^2}

Solving for the charge-to-mass ratio and substituting v=E/Bv = E/B:

qm=2yv2EL2=2yEB2L2\frac{q}{m} = \frac{2 y v^2}{E L^2} = \frac{2 y E}{B^2 L^2}

His value: q/m1.76×1011q/m \approx 1.76 \times 10^{11} C/kg.

Vertical deflection y versus deflecting field E for a velocity-selected cathode-ray beam A straight line through the origin rising to the right, showing that the deflection y of a velocity selected electron beam is directly proportional to the deflecting field E. Four data points sit on the line. The gradient equals q L squared over 2 m v squared. field E (×10⁴ V m⁻¹) deflection y (mm) 12345 481216 gradient = qL² ⁄ (2mv²) y ∝ E: a line through the origin.

What Thomson learned

The measured q/mq/m for cathode rays is about 1800 times larger than for the lightest known ions (hydrogen). Two interpretations were possible: cathode-ray particles either have much larger charge or much smaller mass than hydrogen ions. The same ratio was obtained from any cathode metal (aluminium, platinum, iron), so the particles were a universal constituent. Charge measurements (later refined by Millikan) confirmed the small-mass interpretation.

Thomson concluded:

  • Cathode rays are streams of negatively charged particles.
  • These particles (electrons) are much lighter than any atom.
  • They are present in every kind of matter.

The electron was the first known sub-atomic particle, and the result implied that atoms have internal structure.

Thomson's plum-pudding model

If atoms are electrically neutral and contain negatively charged electrons, they must also contain positive charge. With no clearer picture available, Thomson proposed that atoms consist of a diffuse positively charged sphere with electrons embedded in it like plums in a pudding (or raisins in a bun). The model:

  • explained neutrality (total charge cancels),
  • accommodated the small mass of the electron compared to the atom,
  • predicted that atoms should respond to applied fields in simple ways.

The plum-pudding model survived only until 1909, when Geiger and Marsden's gold foil experiment (under Rutherford) revealed that the positive charge and almost all the mass of the atom are concentrated in a tiny central nucleus.

Examples in context

Example 1. Powerhouse Museum's heritage Crookes tube demo. A Sydney Powerhouse Museum demonstration uses an evacuated tube at V=25,000 VV = 25{,}000 \text{ V}. Electrons emerge from the cathode with KE=eV=1.6×1019×25,000=4.0×1015 JKE = eV = 1.6 \times 10^{-19} \times 25{,}000 = 4.0 \times 10^{-15} \text{ J}. Using KE=12mv2KE = \tfrac{1}{2} m v^2, v=2KE/m=2×4.0×1015/9.11×1031=9.4×107 m/s=0.31cv = \sqrt{2 KE / m} = \sqrt{2 \times 4.0 \times 10^{-15} / 9.11 \times 10^{-31}} = 9.4 \times 10^7 \text{ m/s} = 0.31 c. (Relativistic correction γ1.05\gamma \approx 1.05 adds only 5%5\%.) The cathode-ray beam casts a sharp shadow of a Maltese-cross obstacle on the fluorescent screen, demonstrating straight-line travel and that the rays carry momentum (the cross is deflected magnetically).

Example 2. Replicating Thomson's e/me/m measurement at UNSW. Crossed fields method: an electron beam passes through E=3.0×104 V/mE = 3.0 \times 10^4 \text{ V/m} and B=1.5×103 TB = 1.5 \times 10^{-3} \text{ T}, balanced so the beam is undeflected. Then v=E/B=3.0×104/1.5×103=2.0×107 m/sv = E/B = 3.0 \times 10^4 / 1.5 \times 10^{-3} = 2.0 \times 10^7 \text{ m/s}. With only BB acting, the beam curves with radius r=mv/(eB)r = m v / (e B), so e/m=v/(rB)e/m = v/(rB). For an observed r=0.075 mr = 0.075 \text{ m}: e/m=2.0×107/(0.075×1.5×103)=1.78×1011 C/kge/m = 2.0 \times 10^7 / (0.075 \times 1.5 \times 10^{-3}) = 1.78 \times 10^{11} \text{ C/kg}, within 1%1\% of the accepted 1.76×1011 C/kg1.76 \times 10^{11} \text{ C/kg} Thomson reported in 1897.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC5 marksDescribe how Thomson used crossed electric and magnetic fields to measure the charge-to-mass ratio of the electron. Include the key equations and the role of each field.
Show worked answer →

Thomson accelerated cathode rays through a potential difference and passed them between two parallel plates in a region with both an electric field EE (between the plates) and a magnetic field BB (perpendicular to EE and to the beam direction).

Step 1: balance the two forces so that the beam goes through undeflected. The electric force on an electron is qEqE; the magnetic force is qvBqvB. Setting them equal gives the beam velocity:

v=E/Bv = E/B.

This velocity selector lets him measure vv from the field strengths alone.

Step 2: switch off the magnetic field. The beam is now deflected by EE only. The vertical deflection yy over a horizontal length LL inside the field, with no force after, gives:

y=12at2=12qEm(Lv)2y = \frac{1}{2} a t^2 = \frac{1}{2} \frac{qE}{m} \left( \frac{L}{v} \right)^2.

Solving for the charge-to-mass ratio:

qm=2yv2EL2=2yEB2L2\frac{q}{m} = \frac{2 y v^2}{E L^2} = \frac{2 y E}{B^2 L^2}.

Thomson obtained q/m1.76×1011q/m \approx 1.76 \times 10^{11} C/kg, far larger than the corresponding value for hydrogen ions. He concluded cathode rays are charged particles much lighter than the lightest atom.

Markers reward both fields and their roles, the velocity selector, the deflection equation, and the conclusion (small mass relative to atoms).

2019 HSC3 marksOutline three observations that established the particle nature of cathode rays, and one observation that ruled against them being electromagnetic waves.
Show worked answer →

Three observations supporting particle nature:

  1. Cathode rays are deflected by electric fields, in a direction consistent with them carrying negative charge. Electromagnetic waves are uncharged and would not deflect.

  2. Cathode rays are deflected by magnetic fields. They follow a curved path through the field, as expected for moving charged particles experiencing qv×Bqv \times B.

  3. A small object placed in the beam casts a sharp shadow, consistent with particles travelling in straight lines from the cathode. Cathode rays also turn a small paddle wheel placed in their path, transferring momentum as particles do.

Observation against electromagnetic waves: the rays are deflected by electric fields. Electromagnetic waves carry no net charge and are not deflected by static electric fields. (Equivalently, the q/mq/m measurement made cathode rays definite particles.)

Markers reward three particle-nature observations and one wave-incompatible observation, with clear reasoning.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksState two observations from cathode ray tube experiments that show cathode rays carry negative charge, and name the SI unit of the charge-to-mass ratio q/mq/m.
Show worked solution →

Two observations. (1) Cathode rays are deflected towards the positive plate in an electric field, the direction expected for negative charge. (2) In a magnetic field, cathode rays curve in the direction given by F=qv×B\vec{F} = q\vec{v} \times \vec{B} for a negative charge (opposite to the direction a positive charge would curve).

Unit. q/mq/m is measured in coulombs per kilogram, Ckg1\text{C}\,\text{kg}^{-1}.

Marks: one for any two correct deflection-based observations naming the negative-charge direction, one for the correct unit Ckg1\text{C}\,\text{kg}^{-1}.

foundation3 marksIn Thomson's apparatus, an electron passes through a velocity selector with E=1.8×104 V/mE = 1.8 \times 10^4\ \text{V/m} and B=6.0×104 TB = 6.0 \times 10^{-4}\ \text{T}, arranged so the electric and magnetic forces oppose. Calculate the speed of an electron that travels through undeflected, and state why this speed does not depend on the electron's charge or mass.
Show worked solution →

At balance the electric and magnetic forces are equal: qE=qvBqE = qvB, so v=EBv = \dfrac{E}{B}.

v=1.8×1046.0×104=3.0×107 m/sv = \dfrac{1.8 \times 10^4}{6.0 \times 10^{-4}} = 3.0 \times 10^7\ \text{m/s}.

The charge qq cancels from both sides of qE=qvBqE = qvB, so the selected speed depends only on the field strengths EE and BB, not on the particle's charge or mass. Any particle (of any charge or mass) at this exact speed passes through undeflected.

Marks: one for the force-balance equation qE=qvBqE = qvB, one for v=3.0×107 m/sv = 3.0 \times 10^7\ \text{m/s} with the unit, one for explaining that qq cancels so v=E/Bv = E/B is independent of charge and mass.

foundation2 marksState Thomson's plum-pudding model of the atom in one or two sentences, and give one feature of the atom it correctly explained.
Show worked solution →

The model. Thomson proposed that the atom is a diffuse sphere of positive charge with negatively charged electrons embedded inside it, like plums in a pudding.

Correct feature. It explained the overall electrical neutrality of the atom (the negative electrons cancel the positive charge spread through the sphere).

Marks: one for a correct description of the model (diffuse positive sphere, embedded electrons), one for a correct explained feature (neutrality, or the small mass of the electron relative to the atom).

core4 marksAn electron travelling at 2.5×107 m/s2.5 \times 10^7\ \text{m/s} enters a region of length L=0.045 mL = 0.045\ \text{m} between deflecting plates producing a uniform field E=1.2×104 V/mE = 1.2 \times 10^4\ \text{V/m} (the magnetic field is switched off). Using e/m=1.759×1011 C/kge/m = 1.759 \times 10^{11}\ \text{C/kg}, calculate (a) the acceleration of the electron between the plates and (b) its vertical deflection yy on leaving the field region.
Show worked solution →

(a) Acceleration. a=eEm=(em)E=(1.759×1011)(1.2×104)=2.11×1015 m/s2a = \dfrac{eE}{m} = \left(\dfrac{e}{m}\right)E = (1.759 \times 10^{11})(1.2 \times 10^4) = 2.11 \times 10^{15}\ \text{m/s}^2.

(b) Deflection. Time in the field: t=Lv=0.0452.5×107=1.80×109 st = \dfrac{L}{v} = \dfrac{0.045}{2.5 \times 10^7} = 1.80 \times 10^{-9}\ \text{s}.

y=12at2=0.5×(2.11×1015)×(1.80×109)2=3.4×103 m=3.4 mmy = \tfrac{1}{2}at^2 = 0.5 \times (2.11 \times 10^{15}) \times (1.80 \times 10^{-9})^2 = 3.4 \times 10^{-3}\ \text{m} = 3.4\ \text{mm}.

Marks: one for a=(e/m)Ea = (e/m)E with substitution, one for a=2.11×1015 m/s2a = 2.11 \times 10^{15}\ \text{m/s}^2, one for t=L/vt = L/v and y=12at2y = \tfrac{1}{2}at^2 set up, one for y=3.4 mmy = 3.4\ \text{mm} with correct unit.

core4 marksThe figure shows the vertical deflection yy measured on Thomson's screen as a function of the deflecting field EE, for cathode rays velocity-selected to v=2.0×107 m/sv = 2.0 \times 10^7\ \text{m/s} over a plate length L=0.040 mL = 0.040\ \text{m} (magnetic field off while yy is measured). **(a)** Describe the relationship shown. **(b)** Using the points at E=1.0×104 V/mE = 1.0 \times 10^4\ \text{V/m} and E=4.0×104 V/mE = 4.0 \times 10^4\ \text{V/m}, calculate the gradient. **(c)** Use the gradient to determine the experimental value of q/mq/m, and compare it with the accepted value 1.759×1011 C/kg1.759 \times 10^{11}\ \text{C/kg}.
Show worked solution →

(a) The graph is a straight line through the origin, so yy is directly proportional to EE (yEy \propto E), consistent with y=qL22mv2Ey = \dfrac{qL^2}{2mv^2}E at fixed LL and vv.

(b) Gradient =ΔyΔE=(14.13.5)×103 m(4.01.0)×104 V/m=1.06×1023.0×104=3.53×107 m per V/m= \dfrac{\Delta y}{\Delta E} = \dfrac{(14.1 - 3.5) \times 10^{-3}\ \text{m}}{(4.0 - 1.0) \times 10^{4}\ \text{V/m}} = \dfrac{1.06 \times 10^{-2}}{3.0 \times 10^{4}} = 3.53 \times 10^{-7}\ \text{m per V/m}.

(c) The gradient equals qL22mv2\dfrac{qL^2}{2mv^2}, so qm=2v2×gradientL2=2(2.0×107)2(3.53×107)(0.040)2=1.77×1011 C/kg\dfrac{q}{m} = \dfrac{2v^2 \times \text{gradient}}{L^2} = \dfrac{2(2.0 \times 10^7)^2 (3.53 \times 10^{-7})}{(0.040)^2} = 1.77 \times 10^{11}\ \text{C/kg}.

This is within about 1%1\% of the accepted value 1.759×1011 C/kg1.759 \times 10^{11}\ \text{C/kg}, well within the precision expected of a field-and-ruler measurement.

Marks: one for identifying direct proportionality (line through the origin, yEy \propto E), one for a correct gradient with unit (m per V/m\text{m per V/m}), one for equating the gradient to qL2/(2mv2)qL^2/(2mv^2) and rearranging for q/mq/m, one for q/m1.77×1011 C/kgq/m \approx 1.77 \times 10^{11}\ \text{C/kg} with a sound comparison to the accepted value.

exam6 marksAnalyse how Thomson's crossed-field experiment provided evidence for both the existence of the electron and its properties, and explain why the plum-pudding model was the natural model to propose immediately afterwards.
Show worked solution →

Band-6 plan. (1) State what the crossed-field method measures and how (velocity selector, then deflection). (2) State the numerical result and what made it significant (universal, ~1800x larger than for hydrogen ions). (3) Explain the two interpretations (large charge vs small mass) and how the same-ratio-for-every-metal result supported "new particle" over "special ion". (4) Link directly to why a diffuse positive sphere with embedded electrons was the simplest model consistent with neutrality and a light, universal negative constituent.

Model answer. Thomson first balanced the electric force qEqE against the magnetic force qvBqvB on the beam so that it passed through undeflected, giving the beam speed v=E/Bv = E/B independent of the particle's charge or mass. He then switched off the magnetic field and measured the deflection yy produced by EE alone over the plate length LL, giving q/m=2yE/(B2L2)q/m = 2yE/(B^2L^2). This procedure measures a ratio without needing to know qq or mm separately, which is what makes it possible without first knowing what the particle is.

The measured value, q/m1.76×1011 C/kgq/m \approx 1.76 \times 10^{11}\ \text{C/kg}, was the same regardless of the cathode metal used (aluminium, platinum, iron), which was strong evidence that cathode rays are a single, universal constituent of matter rather than metal-specific ions. Comparing this ratio with the already-known q/mq/m for the hydrogen ion showed it was roughly 1800 times larger. Since later experiments (Millikan) showed the charge was comparable in magnitude to the elementary charge on an ion, the huge ratio meant the particle's mass, not its charge, was the unusual quantity: cathode-ray particles (electrons) are far lighter than any atom.

This directly motivated the plum-pudding model. If atoms are neutral overall but contain very light, negatively charged electrons as a universal constituent, the remaining positive charge and most of the mass had to be accounted for by something else, and with no evidence yet of a concentrated nucleus, the simplest consistent picture was a diffuse sphere of positive charge (carrying the mass and positive charge) with the light electrons embedded inside it, so the atom's total charge cancelled. The model was a direct, minimal extension of what the crossed-field result had actually shown, not a separate speculative leap.

Marker's note: the top band explains the physics of the crossed-field method (why the velocity selector works, why deflection then gives q/mq/m), correctly interprets the large ratio as "small mass" rather than "large charge" (using the universality across metals as the key evidence), and shows the logical chain from that conclusion to the plum-pudding model rather than stating the model in isolation.

exam7 marksEvaluate the extent to which cathode ray tube experiments and Thomson's crossed-field measurement together resolved the 19th century debate about the nature of cathode rays, and assess the limitations of the picture of the atom this left in 1897.
Show worked solution →

Band-6 plan. Thesis: the combined evidence decisively resolved the particle-vs-wave debate but left the internal structure of the atom only partly resolved. Structure: (1) summarise the wave-vs-particle debate and the key deciding evidence (electric-field deflection). (2) explain what the crossed-field q/mq/m measurement added beyond simple deflection (a precise, reproducible, universal quantity). (3) evaluate the limits: no measurement of charge or mass separately, no model of the positive charge, no explanation of atomic stability. (4) reach a judgement.

Model answer. Before 1897 there were two camps: continental physicists who treated cathode rays as a wave phenomenon in the aether, and British physicists who argued for charged particles. The decisive discriminating evidence was deflection by a static electric field. An electromagnetic wave carries no net charge and is not deflected by a static E\vec{E}, whereas cathode rays deflected consistently towards the positive plate. Combined with magnetic deflection, straight-line shadows and the turning of a paddle wheel (showing momentum transfer), this evidence made the particle interpretation essentially unavoidable, resolving the debate.

Thomson's crossed-field measurement went further than qualitative deflection: by balancing qEqE against qvBqvB to select a speed, then measuring the deflection due to EE alone, he obtained a precise, reproducible number, q/m1.76×1011 C/kgq/m \approx 1.76 \times 10^{11}\ \text{C/kg}, independent of the assumed values of qq and mm individually. Crucially, this ratio was the same for every cathode material tested, which is strong evidence that cathode rays are a single universal particle (the electron) rather than an ionised form of whatever gas or metal was used, a claim that simple deflection observations alone could not establish.

However, the resolution was incomplete. The crossed-field method measures only the ratio q/mq/m; Thomson could not separate charge from mass without an independent measurement, which had to wait for Millikan's oil-drop experiment. Nor did the experiment say anything about the positive charge needed to keep atoms neutral: Thomson's plum-pudding model, a diffuse positive sphere with embedded electrons, was a reasonable but essentially unconstrained guess, since no experiment had yet probed how the positive charge was distributed. It also gave no account of atomic stability or of what would later turn out to be a concentrated nucleus, gaps only closed by Geiger, Marsden and Rutherford in 1909-1911.

On balance, cathode ray tube experiments and Thomson's measurement fully resolved the particle-versus-wave question and established the electron as a real, universal, very light charged particle, but they left the arrangement of positive charge in the atom essentially unknown, a limitation the plum-pudding model exposed rather than solved.

Marker's note: the top band separates what was resolved (particle nature, universality, the q/mq/m value) from what was not (separate qq and mm, the structure of the positive charge, atomic stability), and reaches an explicit, weighed judgement rather than simply narrating the experiments. Naming Millikan and Rutherford/Geiger-Marsden as the experiments that closed the remaining gaps is rewarded.

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