Determination of the Rydberg constant from the Balmer series

Theory

In Bohr's model, refined by the reduced-mass correction, the hydrogen terms are \(E_n = -hcR_H/n^2\). Transitions to \(n = 2\) form the Balmer series:

$$ \frac{1}{\lambda} = R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right),\quad n = 3,4,5,\dots \qquad R_H = \frac{R_\infty}{1+m_e/m_p} = 1.09677583\times10^{7}\ \text{m}^{-1}$$

A grating with spacing \(d\), used at normal incidence, sends wavelength λ to angles θ satisfying

$$ d\sin\theta = m\lambda .$$

Reading the telescope on both sides of the direct image gives 2θ, which cancels the unknown zero of the circle and, to first order, a small error in setting the grating normal.

Apparatus

• Spectrometer with a divided circle and two verniers (least count 1′)
• Plane transmission grating (300/600/1200 lines per mm)
• Hydrogen discharge tube with power supply, and a mercury lamp for calibration
• Spirit level and reading lens

Procedure

1. Level the instrument, focus the eyepiece on the cross-wires and the telescope at infinity (Schuster's method), and make the slit narrow and vertical.
2. Set the grating normal to the collimator. The direct (white) image lies at φ = 0.
3. Calibration: select the Hg lamp. Centre the green 546.07 nm line on the cross-wire on the left, record both verniers, then repeat on the right.
4. Switch to the hydrogen lamp. For each Balmer line (red Hα, blue-green Hβ, violet Hγ, deep-violet Hδ) record the left and right readings in first order, and in second order if visible.
5. Compute 2θ, then θ and λ = d sin θ / m. Plot 1/λ against (1/4 − 1/n²): a straight line through the origin with slope \(R_H\).

Analysis details

• Uncertainty: each vernier reading has ±½′ reading error, so σ(θ) ≈ 0.35′. This propagates to \(\sigma_\lambda = (d\cos\theta/m)\,\sigma_\theta\), and the fit is weighted accordingly.
• Air vs vacuum: the grating measures wavelength in air. The Rydberg formula refers to vacuum, and \(\lambda_{vac} = n_{air}\lambda_{air}\) with \(n_{air} - 1 \approx 2.77\times10^{-4}\). Untick the box to see the 0.03% bias this causes.
• Ionisation energy: \(E_I = hcR_H \approx 13.598\) eV.
• Deuterium: the larger nuclear mass increases \(R\) by 0.027%, shifting Hα by 0.18 nm toward shorter wavelength (Urey, 1932). Try to resolve it in second order with the 1200 lines/mm grating.
• Fine structure (Dirac theory) shifts the real line centroids by about 0.01 nm from the Bohr values. This is below the resolution of this method.

Precautions & sources of error

• The slit must be narrow; centre each line, not its edge, on the cross-wire.
• Always read both verniers; this removes the eccentricity error of the circle.
• Clamp the telescope and use the tangent screw for the final setting; avoid backlash by always approaching from the same side.
• Hδ is faint, and Hε lies near the UV limit of the eye.

Viva questions

Why measure on both sides of the direct image?

The circle's zero position is unknown. The difference of two readings gives 2θ directly, and averaging left and right cancels first-order errors from incidence that isn't quite normal.

Why read both verniers?

If the axis of rotation is not exactly at the centre of the scale (eccentricity), one vernier reads too high and the other too low by the same amount. The mean is correct.

Why is R_H smaller than R∞?

The proton is not infinitely heavy. The electron and proton orbit their common centre of mass, so the electron mass is replaced by the reduced mass μ = m_e m_p /(m_e + m_p).

What limits the resolving power?

R = λ/Δλ = mN, where N is the number of illuminated rulings. Higher orders and wider illuminated gratings resolve finer detail.