Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

WHAT is it?

WHY does it look like $1/n^2$ ? — Derivation from scratch

We derive it from the Bohr model so no step is magic.

Step 1 — Energy of a Bohr level. Bohr's quantisation gives the energy of level $n$ in hydrogen. Because both the electron and the proton orbit their common centre of mass, the correct mass to use is the reduced mass $\mu = \dfrac{m_e m_p}{m_e + m_p}$ , not the bare electron mass $m_e$ : $E_n = -\frac{13.6\ \text{eV}}{n^2} = -\frac{\mu e^4}{8\varepsilon_0^2 h^2}\cdot\frac{1}{n^2}$ Why this step? The electron is bound (negative energy), and the $1/n^2$ comes from balancing Coulomb attraction against the quantised angular momentum $L=n\hbar$ . The energy levels get closer together as $n$ grows. Using $\mu$ instead of $m_e$ accounts for the finite (not infinite) proton mass.

Step 2 — Energy of the emitted photon. When the electron drops from $n_2 \to n_1$ , energy is conserved: the lost atomic energy becomes one photon. $\Delta E = E_{n_2} - E_{n_1} = -\frac{\mu e^4}{8\varepsilon_0^2 h^2}\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right)$ The photon energy is the amount released $= E_{n_2}-E_{n_1}$ but taken positive: $E_{\text{photon}} = E_{n_2}-E_{n_1} = \frac{\mu e^4}{8\varepsilon_0^2 h^2}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ Why this step? Since $n_2>n_1$ , $\frac{1}{n_1^2}>\frac{1}{n_2^2}$ , so the bracket is positive → photon energy is positive. Good.

Step 3 — Convert photon energy to wavelength. A photon has $E_{\text{photon}} = h\nu = \dfrac{hc}{\lambda}$ . $\frac{hc}{\lambda} = \frac{\mu e^4}{8\varepsilon_0^2 h^2}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ Why this step? We want a formula in $\lambda$ , so replace $E$ by $hc/\lambda$ .

Step 4 — Solve for $1/\lambda$ . Divide both sides by $hc$ : $\boxed{\;\frac{1}{\lambda} = \underbrace{\frac{\mu e^4}{8\varepsilon_0^2 h^3 c}}_{\displaystyle R_H}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\;}$

Formula The Rydberg constant is NOT arbitrary

The experimental Rydberg constant for hydrogen uses the reduced mass: $R_H = \frac{\mu\, e^4}{8\varepsilon_0^2 h^3 c}, \qquad \mu = \frac{m_e m_p}{m_e+m_p}$ If we pretend the proton is infinitely heavy ( $m_p\to\infty$ , so $\mu\to m_e$ ), we get the idealised constant $R_\infty = \frac{m_e\, e^4}{8\varepsilon_0^2 h^3 c} \approx 1.0974\times10^{7}\ \text{m}^{-1}.$ Since $\mu = m_e\,\dfrac{m_p}{m_e+m_p} \approx 0.99946\,m_e$ , we get $R_H = R_\infty\cdot\dfrac{\mu}{m_e} \approx 1.0968\times10^{7}\ \text{m}^{-1}$ — very slightly smaller than $R_\infty$ . This tiny shift is the whole point of Bohr's triumph refined: a number measured in spectra comes out of pure theory, and even the finite proton mass shows up.

Spectral series (the crowding staircase)

$n_1$ fixes the series (where the electron lands); $n_2$ fixes the line within it.

Series	$n_1$	Region
Lyman	1	Ultraviolet
Balmer	2	Visible
Paschen	3	Infrared
Brackett	4	Infrared
Pfund	5	Far Infrared

Figure — Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

Worked Examples

Worked example 1. The red H-alpha line (Balmer

$3\to2$ ) Find $\lambda$ for $n_2=3 \to n_1=2$ . $\frac{1}{\lambda} = R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) = 1.097\times10^7\left(\frac14-\frac19\right)$ Why this step? Balmer means it lands on $n_1=2$ ; the line comes from $n_2=3$ . $\frac14-\frac19 = \frac{9-4}{36}=\frac{5}{36}=0.1389$ $\frac{1}{\lambda}=1.097\times10^7\times0.1389 = 1.524\times10^6\ \text{m}^{-1}$ $\lambda = \frac{1}{1.524\times10^6} = 6.56\times10^{-7}\ \text{m} = 656\ \text{nm} \;(\text{red})\;\checkmark$

Common Mistakes (Steel-manned)

Active Recall

Recall Forecast-then-Verify: which series is visible, and why?

Forecast first, then open. Answer: The Balmer series ( $n_1=2$ ). Only its lines fall in the 380–700 nm visible window because the energy gaps landing on $n=2$ happen to match visible photon energies. Lyman ( $n_1=1$ ) has bigger gaps → UV; Paschen and beyond have smaller gaps → IR.

Recall Explain to a 12-year-old (Feynman)

Imagine an electron sitting on a staircase inside the atom. The steps get squished closer together the higher you go. When the electron slides down to a lower step, it must throw away energy — and it throws it out as a flash of coloured light. A big drop = a big flash = bluer light; a small drop = a small flash = redder light. The Rydberg formula is just the rulebook that tells you exactly what colour comes out for each slide down the staircase, using only the step numbers you start and end on.

Flashcards

What does

n_1

represent in the Rydberg formula?

The quantum number of the lower (final, smaller) energy level; the series terminus.

Why must

n_2 > n_1

So the bracket

(1/n_1^2 - 1/n_2^2)

stays positive, giving a positive wavenumber and a real, positive wavelength.

What is the value and units of

R_H

\approx 1.097\times10^{7}\ \text{m}^{-1}

(Rydberg constant for hydrogen).

Which mass appears in the true Rydberg constant, and why?

The reduced mass

\mu = m_e m_p/(m_e+m_p)

, because both electron and proton orbit their common centre of mass.

What is

R_\infty

versus

R_H

R_\infty = m_e e^4/(8\varepsilon_0^2 h^3 c)

is the infinite-nuclear-mass idealisation;

R_H = R_\infty\,\mu/m_e

is slightly smaller for real hydrogen.

Which spectral series lies in the visible region and what is

n_1

Balmer series,

n_1 = 2

What is

1/\lambda

called and what is it proportional to?

The wavenumber

\bar\nu

; proportional to the photon energy

E = hc/\lambda

How is the Rydberg formula modified for a hydrogen-like ion of charge Z?

1/\lambda = R Z^2 (1/n_1^2 - 1/n_2^2)

(with its own reduced mass).

What transition gives the series limit (shortest λ) of a series?

n_2 \to \infty

, giving

1/\lambda = R_H/n_1^2

Derive

R_H

symbolically from Bohr energies.

R_H = \mu e^4 / (8\varepsilon_0^2 h^3 c)

, from equating

hc/\lambda

to the Bohr energy gap.

Mnemonic

Connections

Bohr Model of the Atom — supplies $E_n = -13.6/n^2$ eV that the whole derivation rests on.
Reduced Mass in Two-Body Systems — why $\mu$ replaces $m_e$ in the real Rydberg constant.
Energy of a Photon E = hc over lambda — the bridge from energy gap to wavelength.
Hydrogen Spectrum and Spectral Series — the experimental lines this formula predicts.
Quantisation of Angular Momentum — origin of the $1/n^2$ energy dependence.
Ionisation Energy — the series limit ( $n_2\to\infty$ ) equals ionising the ground state.
Hydrogen-like Ions and Z-scaling — generalisation with the $Z^2$ factor.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, hydrogen atom ke andar electron alag-alag "orbits" ya energy levels par baith sakta hai, jaise seedhi ki steps. Bohr ne bataya ki har level ki energy $E_n = -13.6/n^2$ eV hoti hai — matlab upar jaate jaate steps ek-doosre ke bahut kareeb aa jaate hain. Jab electron upar wale level ( $n_2$ ) se neeche wale level ( $n_1$ ) par girta hai, to jo extra energy hoti hai wo ek photon (light ka packet) ban ke bahar nikal jaati hai. Bas yahi energy conservation ko wavelength ki bhaasha mein likh do, to Rydberg formula ban jaata hai: $1/\lambda = R_H(1/n_1^2 - 1/n_2^2)$ .

Ek important detail: proton infinitely heavy nahi hai, isliye electron aur proton dono apne common centre of mass ke around ghoomte hain. Isliye formula mein bare electron mass $m_e$ nahi, balki reduced mass $\mu = m_e m_p/(m_e+m_p)$ aata hai. Iski wajah se real hydrogen ka $R_H$ idealised $R_\infty$ se thoda sa (lagbhag 0.05%) chhota hota hai. Exam ke numbers mein ye choti si baat hai, par concept ke liye yaad rakho — "reduce, don't reuse".

Yaad rakho — $n_1$ hamesha chhota (neeche wala) level hai aur $n_2$ bada. Isko ulta mat karna, warna $\lambda$ negative aa jaayega. $1/\lambda$ ko wavenumber bolte hain aur ye seedha energy ke proportional hai. Series ka funda simple hai: $n_1$ decide karta hai kaun si series ( $n_1=1$ Lyman-UV, $n_1=2$ Balmer-visible, $n_1=3$ Paschen-IR), aur $n_2$ us series ke andar kaun si line. Balmer hi humein aankhon se dikhta hai. Aur bonus: $n_2 \to \infty$ karo to milta hai series limit, jo actually ionisation hai.

Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

WHAT is it?

WHY does it look like $1/n^2$ ? — Derivation from scratch

Spectral series (the crowding staircase)

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Flashcards

Mnemonic

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Atomic Structure (Classical)

Connections