University Physics IV · Angular Momentum and Spin · 9.5

The Stern–Gerlach Experiment

This is where "μ_z 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.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Send a magnetic moment through a field that varies across the beam and the field grades into a force: F_z = μ_z(∂B_z/∂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 μ_z 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 μ_z alone: the ±μ_B 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.

Simple definition: The Stern–Gerlach experiment passes a beam of neutral atoms through a strongly inhomogeneous magnetic field, so each atom deflects in proportion to μ_z; 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 μ_z = ±μ_B traces about 0.3 mm apart at the exit — two stripes, and nothing at the undeflected centre.

Force in an inhomogeneous fieldF_z = μ_z (∂B_z/∂z)

Turns an invisible internal moment into a transverse deflection you can catch on a plate.

μ_z in J T⁻¹, gradient in T m⁻¹, force in N; a uniform field, however strong, gives torque and precession only

Deflection through magnet and driftz = (μ_z B′ L / m v²)(L/2 + D)

Run it forward to predict the split; run it backward on a measured split to extract μ_z.

B′ = ∂B_z/∂z (T m⁻¹); L magnet length and D drift length (m); m atomic mass (kg), v speed (m s⁻¹)

Quantised projectionμ_z = −g mⱼ μ_B, mⱼ = −j … +j

Integer l always gives an odd count with an undeflected centre; an even count forces half-integer j.

μ_B = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹ = 5.788 × 10⁻⁵ eV T⁻¹; number of traces = 2j + 1

Silver'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 ~10³ times smaller than μ_B

Oven beam speedsvₘₚ = √(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), k = 1.38 × 10⁻²³ J K⁻¹; silver at 1300 K gives v ≈ 550 m s⁻¹

A 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, about which the moment precesses at the Larmor frequency while the atom flies straight on. Force needs a gradient: F_z = μ_z(∂B_z/∂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 μ_y swing round of order a million times during the transit, average to zero, and deflect nothing. Put in μ_z = μ_B 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.

Classical band versus quantised stripes

What lands on the plate is a histogram of μ_z. Classically an oven randomises orientation, so μ_z = 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.

From force to millimetres on the plate

Inside the magnet the atom is a projectile: constant transverse acceleration a = μ_zB′/m ≈ 7.3 × 10⁴ m s⁻² for μ_z = μ_B — 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 = (μ_zB′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.

Silver 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 comes with the electron mass in μ_B replaced by a nuclear mass — three orders of magnitude smaller, 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.

What two traces prove, and the g they cannot

Be strict about what the plate measures: μ_z, nothing else, and it came out close to ±μ_B. Since μ_z = −g m μ_B, 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 ±μ_B 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Gradient dBz/dz1400 T/m

Atom speed v550 m/s

Spin g factor2.0

Slow the beam from 550 to 390 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 ±μ_B, which s = 1/2 can only deliver with g near 2.

|μz| = (g/2) μB1.00 μB

FORCE Fz130 10⁻²² N

SPLIT AT MAGNET EXIT0.29 mm

SPLIT AT PLATE1.97 mm

Live interpretation|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

Catch the common trap

Explain before calculating.

Misconception check

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 = −μ_zB_z gives F_z = μ_z ∂B_z/∂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 ∂B_z/∂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 μ_z matters: μₓ and μ_y average to zero over each turn, so their gradients deflect nothing.

A beam of ground-state hydrogen atoms (1s, so l = 0) passes down a Stern–Gerlach magnet. What lands on the plate?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA silver atom (m = 1.79 × 10⁻²⁵ kg) with μ_z = μ_B = 9.27 × 10⁻²⁴ J T⁻¹ enters a Stern–Gerlach magnet whose gradient is ∂B_z/∂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.

F_z = μ_z (∂B_z/∂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: F_z/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 = F_z/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.

AnswerF_z = 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.

MediumSilver 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 μ_z = ±μ_B. Find the beam speed and the separation between the two beams at the magnet exit.

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 μ_z = +μ_B: a = μ_B 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.

Answerv ≈ 550 m s⁻¹; each beam moves 0.15 mm off axis, so the split at the exit is about 0.3 mm.

HardIn 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 |μ_z|. 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.

Each trace sits z = 1.0 mm off axis. Deflection inside the magnet plus straight-line drift gives z = (μ_z B′ L / m v²)(L/2 + D).
Lever arm: L/2 + D = 0.0175 + 0.100 = 0.1175 m.
Solve for the moment: μ_z = 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: μ_z = 9.4 × 10⁻²⁴ J T⁻¹ = 1.01 μ_B.
With |μ_z| = gₛ |mₛ| μ_B 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|μ_z| ≈ 9.4 × 10⁻²⁴ J T⁻¹ ≈ 1.0 μ_B, 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.