
Physics IV course sequence · Interactive lesson
The Stern–Gerlach experiment
This is where "μz is quantised" stops being algebra and becomes silver on a glass plate. Get the gradient force, the oven kinematics and silver's electron count right and you can predict two stripes a fraction of a millimetre apart — and say exactly which part of the spin story those stripes do not prove.
UPIV-09.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Build the model
The apparatus measures μz — and μz came out in exactly two values.
The key idea
The Stern–Gerlach experiment passes a beam of neutral atoms through a strongly inhomogeneous magnetic field, so each atom deflects in proportion to μz; the beam lands in 2j + 1 discrete traces rather than the continuous band a classical moment distribution would leave.
Example. Silver atoms at 550 m s⁻¹ crossing a 3.5 cm magnet with gradient 1.4 × 10³ T m⁻¹ split into μz = ±μB traces about 0.3 mm apart at the exit — two stripes, and nothing at the undeflected centre.
The complete physical model and assumptions
Send a magnetic moment through a field that varies across the beam and the field grades into a force: Fz = μz(∂Bz/∂z), so each atom's deflection reads out one component of its moment. Classically that component is μ cos θ with θ randomised by the oven, so the beam should land as one continuous band. Quantum mechanics says μz is built from a quantised projection — 2j + 1 discrete values — so the beam must land in discrete traces. Stern and Gerlach ran the test in 1922 with silver: an oven beam, a knife-edge pole to make the gradient huge, and a plate that collected two stripes with nothing at the centre. The result is both stronger and weaker than it looks. Stronger, because every integer l predicts an odd count of traces including an undeflected one, so two stripes rule out the orbital ladder wholesale and force a half-integer j. Weaker, because the apparatus measures μz alone: the ±μB it recorded fixes only the product g⋅m, and s = 1/2 reproduces it only if the spin moment carries g ≈ 2 — a number the deflections cannot supply. Stern and Gerlach themselves read the plate as Bohr–Sommerfeld space quantisation with l = 1. The lasting lesson is the cost: a measurement fixes the eigenvalues of the quantity it couples to, and nothing else.
1A gradient, not a field, deflects the atom
A dipole in a magnetic field has energy U = −μ⋅B. In a uniform field that energy is the same everywhere, so there is no net force — only a torque, under which the moment precesses about the field at the Larmor frequency while the atom flies straight on. Force needs a gradient: Fz = μz(∂Bz/∂z). That is why Gerlach machined one pole to a knife edge facing a grooved pole — the geometry buys a gradient around 1.4 × 10³ T m⁻¹ across a beam a slit-width wide. The precession is also why only the z-component matters: μₓ and μy swing round of order a million times during the transit, average to zero, and deflect nothing. Put in μz = μB and the force is 1.3 × 10⁻²⁰ N — feeble, yet 7 × 10³ times the silver atom's weight, and applied to a neutral atom, so there is no Lorentz force qv × B to drown it. That last clause is why the experiment works with atoms and fails with free electrons.
2Classical band versus quantised stripes
What lands on the plate is a histogram of μz. Classically an oven randomises orientation, so μz = μ cos θ with cos θ uniformly distributed: every value between −μ and +μ is equally represented, and the prediction is one continuous band. Quantum mechanics instead allows only discrete projections, so a moment of quantum number j lands in 2j + 1 traces. Run the count for orbital angular momentum: l = 0 gives one undeflected trace, l = 1 gives three, l = 2 gives five — always an odd number, and always with a trace at the centre, because integer l includes m = 0. The plate showed two stripes and an empty centre. That single photograph rules out the classical band and every integer-l ladder at once: whatever splits the beam has 2j + 1 = 2, hence j = 1/2, and half-integers never occur for orbital motion.
3From force to millimetres on the plate
Inside the magnet the atom is a projectile: constant transverse acceleration a = μzB′/m ≈ 7.3 × 10⁴ m s⁻² for μz = μB — about 7400g — held for a time t = L/v. A 550 m s⁻¹ atom spends 64 μs in a 3.5 cm magnet and exits z = ½at² ≈ 0.15 mm off axis, so the two spin states sit about 0.3 mm apart, the scale of the historic traces. A field-free drift D then multiplies the effect through the exit slope: z = (μzB′L/mv²)(L/2 + D), and 10 cm of drift stretches 0.3 mm to about 2 mm. Note what the formula punishes: deflection goes as 1/v², and the oven supplies a Maxwell–Boltzmann spread of speeds, so each trace smears into a streak — fast atoms crowd the centre, slow ones feather outward. The two-lobed 1922 pattern was resolvable because the collimating slit was narrower than the splitting, not because the atoms shared one speed.
4Silver is a spin wearing a heavy neutral coat
Silver was luck as much as design, and the luck is in its electron count. Forty-six of the 47 electrons close into [Kr] 4d¹⁰, and a filled subshell has zero total orbital and zero total spin angular momentum. Everything is carried by the lone 5s electron, and s states have l = 0: the ground term is ²S₁/₂, so silver has no orbital moment to deflect at all. The nuclear moment is real but scales with the proton mass where μB has the electron mass, and silver's is about 10⁴ times smaller than μB — invisible at this resolution. The beam is effectively one free electron spin per atom, chaperoned by 108 u of neutral ballast that kills the Lorentz force. A two-way split of an l = 0 atom is maximally awkward for orbital explanations, which is why the decisive repeat was Phipps and Taylor's 1927 run with ground-state hydrogen: one electron, certainly 1s, certainly l = 0 — and still two traces.
5What two traces prove, and the g they cannot
Be strict about what the plate measures: μz, nothing else, and it came out close to ±μB. Since μz = −g m μB, a deflection fixes only the product g⋅m. Old quantum theory factorised it as g = 1, m = ±1 with m = 0 mysteriously absent — that is the space quantisation Stern and Gerlach announced in 1922, three years before spin was proposed. Modern mechanics factorises the same number as g ≈ 2, mₛ = ±1/2. The deflections cannot tell these apart; what breaks the tie is outside evidence. Spectroscopy gives silver an l = 0 ground state, so orbital m = ±1 is unavailable, and the anomalous Zeeman analysis — later Dirac's equation — supplies gₛ ≈ 2. Only with that imported g does ±μB become s = 1/2. The habit generalises: a measurement returns an eigenvalue of the operator it couples to, and every further conclusion is inference that must be paid for elsewhere.
Key equations
Read the units, assumptions and limits before substituting.
Force in an inhomogeneous field
Fz = μz (∂Bz/∂z)
Turns an invisible internal moment into a transverse deflection you can catch on a plate.
μz in J T⁻¹, gradient in T m⁻¹, force in N; a uniform field, however strong, gives torque and precession onlyDeflection through magnet and drift
z = (μz B′ L / m v²)(L/2 + D)
Run it forward to predict the split; run it backward on a measured split to extract μz.
B′ = ∂Bz/∂z (T m⁻¹); L magnet length and D drift length (m); m atomic mass (kg), v speed (m s⁻¹)Quantised projection
μz = −g mⱼ μB, mⱼ = −j … +j
Integer l always gives an odd count with an undeflected centre; an even count forces half-integer j.
μB = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹ = 5.788 × 10⁻⁵ eV T⁻¹; number of traces = 2j + 1Silver's ground term
[Kr] 4d¹⁰ 5s¹ → ²S₁/₂ (L = 0, S = ½)
The whole deflecting moment is one electron's spin, so the two-way split cannot be orbital.
filled subshells carry zero L and zero S; the nuclear moment is about 10⁴ times smaller than μBOven beam speeds
vₘₚ = √(3kT/m) for the flux-weighted beam
Sets the deflection scale — and the spread of speeds smears each 1/v² trace into a streak.
T oven temperature (K), m atomic mass (kg), k = 1.38 × 10⁻²³ J K⁻¹; silver at 1300 K gives v ≈ 550 m s⁻¹Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Slow the beam from 600 to 425 m s⁻¹ and the split doubles — deflection goes as 1/v². Then drop the g factor from 2 toward 1 at fixed gradient and the traces close to half: the 1922 plate showed ±μB, which s = 1/2 can only deliver with g near 2.
Live results
- |μz| = (g/2) μB
- 1.00 μB
- FORCE Fz
- 130 10⁻²² N
- SPLIT AT MAGNET EXIT
- 0.29 mm
- SPLIT AT PLATE
- 1.97 mm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Gradient ∂Bz/∂z: 1400 T/m; Atom speed v: 550 m/s; Spin g factor: 2.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students reach for "the field pushes the moments apart" and size the effect by B. But a dipole in a uniform field feels zero net force however strong the field: U = −μzBz gives Fz = μz ∂Bz/∂z, and with no gradient there is nothing to differentiate — the field only exerts a torque, and the moment precesses about it at the Larmor frequency while the atom flies on undeflected. That is why Gerlach machined one pole to a knife edge: the point was never a large B but a large ∂Bz/∂z, around 10³ T m⁻¹ across the beam. Doubling the field at fixed gradient changes the precession rate and nothing on the plate; doubling the gradient doubles the split. The same precession is also why only μz matters: μₓ and μy average to zero over each turn, so their gradients deflect nothing.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A silver atom (m = 1.79 × 10⁻²⁵ kg) with μz = μB = 9.27 × 10⁻²⁴ J T⁻¹ enters a Stern–Gerlach magnet whose gradient is ∂Bz/∂z = 1.4 × 10³ T m⁻¹. Find the magnetic force on it, compare it with the atom's weight, and give the transverse acceleration.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Fz = μz (∂Bz/∂z) = 9.27 × 10⁻²⁴ J T⁻¹ × 1.4 × 10³ T m⁻¹ = 1.30 × 10⁻²⁰ N.
- Weight: W = mg = 1.79 × 10⁻²⁵ kg × 9.81 m s⁻² = 1.76 × 10⁻²⁴ N.
- Ratio: Fz/W = 1.30 × 10⁻²⁰ / 1.76 × 10⁻²⁴ ≈ 7.4 × 10³, so gravity barely bends the beam during its ~10⁻⁴ s flight.
- Transverse acceleration: a = Fz/m = 1.30 × 10⁻²⁰ / 1.79 × 10⁻²⁵ = 7.3 × 10⁴ m s⁻², about 7400g — enormous per kilogram, yet it acts only for tens of microseconds.
Answer. Fz = 1.30 × 10⁻²⁰ N, about 7.4 × 10³ times the atom's weight; the atom accelerates sideways at 7.3 × 10⁴ m s⁻².
MediumExample 2
Silver atoms leave an oven at 1300 K, so a typical beam atom moves at v = √(3kT/m). They cross a magnet of length L = 3.5 cm with gradient 1.4 × 10³ T m⁻¹, and μz = ±μB. Find the beam speed and the separation between the two beams at the magnet exit.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Speed: 3kT = 3 × 1.38 × 10⁻²³ × 1300 = 5.38 × 10⁻²⁰ J, so v = √(5.38 × 10⁻²⁰ / 1.79 × 10⁻²⁵) = √(3.01 × 10⁵) ≈ 550 m s⁻¹.
- Time in the magnet: t = L/v = 0.035 / 550 = 6.4 × 10⁻⁵ s.
- Transverse acceleration for μz = +μB: a = μB B′/m = 1.30 × 10⁻²⁰ / 1.79 × 10⁻²⁵ = 7.3 × 10⁴ m s⁻².
- Each beam deflects z = ½at² = 0.5 × 7.3 × 10⁴ × (6.4 × 10⁻⁵)² = 1.5 × 10⁻⁴ m ≈ 0.15 mm.
- The two mₛ states deflect oppositely, so the separation is 2z ≈ 0.3 mm — the scale of the 1922 traces, resolvable only because the collimating slit was narrower still.
Answer. v ≈ 550 m s⁻¹; each beam moves 0.15 mm off axis, so the split at the exit is about 0.3 mm.
HardExample 3
In a modern repeat, the magnet of the last example (L = 3.5 cm, B′ = 1.4 × 10³ T m⁻¹) is followed by a 10 cm field-free drift to the plate, where the two silver traces land 2.0 mm apart. Using v = 550 m s⁻¹, extract |μz|. Then, given that spectroscopy fixes silver's ground state as ²S₁/₂ (j = s = 1/2), find gₛ — and state what the 2.0 mm alone could not have told you.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Each trace sits z = 1.0 mm off axis. Deflection inside the magnet plus straight-line drift gives z = (μz B′ L / m v²)(L/2 + D).
- Lever arm: L/2 + D = 0.0175 + 0.100 = 0.1175 m.
- Solve for the moment: μz = z m v² / [B′ L (L/2 + D)] = (1.0 × 10⁻³ × 1.79 × 10⁻²⁵ × 550²) / (1.4 × 10³ × 0.035 × 0.1175).
- Numerator 5.4 × 10⁻²³, denominator 5.76: μz = 9.4 × 10⁻²⁴ J T⁻¹ = 1.01 μB.
- With |μz| = gₛ |mₛ| μB and mₛ = ±1/2: gₛ = 9.4 × 10⁻²⁴ / (0.5 × 9.27 × 10⁻²⁴) ≈ 2.0.
- The plate alone fixes only the product g⋅m = ±1. Old quantum theory read that as g = 1, m = ±1; it takes j = 1/2 from spectroscopy to turn the same 2.0 mm into gₛ ≈ 2.
Answer. |μz| ≈ 9.4 × 10⁻²⁴ J T⁻¹ ≈ 1.0 μB, so with mₛ = ±1/2, gₛ ≈ 2.0. The split fixes only the product g⋅mⱼ; the half-integer j must be supplied from elsewhere.
Quick check
Select an answer and explain it before checking.
Review & continue
The Stern–Gerlach experiment passes a beam of neutral atoms through a strongly inhomogeneous magnetic field, so each atom deflects in proportion to μz; the beam lands in 2j + 1 discrete traces rather than the continuous band a classical moment distribution would leave.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.