ARCHETYPE: Bohr-Model & Spectral-Line Numericals

The two-line ladder you just drilled is still the working language of astronomy. When astronomers point a spectrograph at a star or a glowing nebula, they read its composition straight off the Balmer and Lyman lines of hydrogen — and they measure the universe's expansion by how far those same $656$ nm and $122$ nm lines have been red-shifted, a calculation that begins with $\lambda = 1240/\Delta E$. The $Z^2$ scaling of example 4 is not a curiosity: hydrogen-like highly charged ions are made routinely in ion traps and tokamak plasmas, and their far-ultraviolet and X-ray lines, sitting exactly where $Z^2$ says they should, diagnose the temperature of fusion plasmas and the solar corona. And the Bohr–Sommerfeld quantum numbers — refined into today's $n$, $\ell$, $m$ and spin — are the labels on every orbital in every chemistry textbook, the scaffold on which the whole periodic table is built. A model that a modern physicist calls "wrong in its picture but right in its numbers" turns out to be right enough to run an industry.

Bookkeeping first, because in a numerical gym the accounting is the physics. Bohr's picture of a one-electron atom — hydrogen, or a bare ion like He⁺ or Li²⁺ that has been stripped down to a single electron — is a nucleus of charge $+Ze$ with one electron circling it on a set of allowed orbits and no others. Two postulates fix everything. First, the electron's angular momentum is quantised in whole steps: $mvr = n\,h/2\pi$, with $n = 1, 2, 3, \dots$ the principal quantum number. Second, while it sits on an allowed orbit it does not radiate — it is a stationary state — and light appears only when the electron jumps from one orbit to another, carrying off the energy difference as a single photon, $\Delta E = hf$. Feed the quantisation rule into Newton's second law for a circular orbit (Coulomb attraction supplying the centripetal force) and three results drop out, one for each thing you can be asked to find. The orbit radius, the orbital speed, and the total energy of the $n$-th level of a hydrogen-like atom of nuclear charge $Ze$ are

where: $n$ — the principal quantum number ($1$ for the ground state, $2, 3, \dots$ for excited states); $Z$ — the nuclear charge number ($1$ for hydrogen, $2$ for He⁺, $3$ for Li²⁺); $0.529$ Å is the Bohr radius $a_0$; $2.19\times10^6$ m/s is the ground-state speed of the electron in hydrogen; and $-13.6$ eV is the ground-state energy of hydrogen. Read the three scalings off the exponents: the orbit swells as $n^2$ and shrinks as $1/Z$, the electron slows as $1/n$ and speeds up as $Z$, and the binding deepens as $Z^2$ and weakens as $1/n^2$. The single most important line is the energy. It is negative — the electron is bound, trapped in the nucleus's well, and you would have to add energy to free it. So $n=1$ (energy $-13.6$ eV for hydrogen) is the deepest, most tightly held rung, the ground state; $n=2$ at $-3.4$ eV is higher up and less bound; and $n=\infty$ at $0$ eV is the top of the well, where the electron is just barely free. Climbing the ladder costs energy; falling down it releases energy as light.

To turn a jump into a wavelength you do not need any new physics — just the second postulate and the photon shortcut. When the electron falls from a higher rung $n_2$ to a lower rung $n_1$, the emitted photon carries the whole gap,

where: $n_1$ — the lower level (smaller $n$, the one landed on); $n_2$ — the upper level (larger $n$, the one started from); $\Delta E$ — the photon energy; $\lambda$ — its wavelength. The bracket is always positive because the lower level has the larger $1/n^2$: this is the arithmetic guard that keeps your energy release positive. For hydrogen's famous red line, the $3\to2$ Balmer-$\alpha$, this gives $\Delta E = 13.6(1/4 - 1/9) = 1.89$ eV and $\lambda = 1240/1.89 = 656$ nm — the wavelength Ångström had measured in 1868 and that Balmer's 1885 formula reproduced, decades before Bohr explained why.

Four tools run the entire family, and every question is one of them solved for whichever letter is hidden. The first is the energy ladder of equation (1), $E_n = -13.6\,Z^2/n^2$ eV. Committing three numbers unlocks most of hydrogen: $E_1 = -13.6$, $E_2 = -3.4$, $E_3 = -1.51$, and thereafter it flattens fast toward $0$. The whole point of the negative sign is that it makes two quantities free at once — the energy of a level and the energy to remove the electron from it. The second tool is the ionisation energy, the energy needed to lift the electron from level $n$ all the way out to $n=\infty$ where $E=0$. That cost is simply the depth of the well at that rung:

where: $E_{\text{ion}}(n)$ — the ionisation energy measured from level $n$. This is where marks bleed away: $13.6$ eV is the ionisation energy of hydrogen only from its ground state. From the first excited state $n=2$ it is just $13.6/4 = 3.4$ eV; for a helium ion (He⁺, $Z=2$) from its ground state it is $13.6\times4 = 54.4$ eV. State the level and the charge before you quote a number. The third tool is the transition-to-wavelength rule of equation (2), $\lambda = 1240/\Delta E$. Because a photon's energy is the difference of two ladder rungs, the same jump written for a hydrogen-like ion of charge $Z$ has every energy multiplied by $Z^2$, so its wavelength is divided by $Z^2$: He⁺ emits hydrogen's $656$ nm red line at $656/4 = 164$ nm, out in the ultraviolet. This $Z^2$-shrink of the whole spectrum is the single most-tested competitive-exam idea in the topic. The fourth tool is the line count. If a gas is excited so that its atoms reach level $n$, the electrons can cascade back down by any route, and the number of distinct photon energies — hence distinct spectral lines — is the number of ways to pick two rungs out of $n$:

where: $N_{\text{lines}}$ — the number of different wavelengths the de-exciting gas can emit. Excite to $n=4$ and you get $4\times3/2 = 6$ lines; to $n=5$, $5\times4/2 = 10$. The classic error is to answer $n-1$ (only the lines that land on the ground state) — that counts one series, not the whole spectrum.

Find an orbit's size, speed or energy. Plug $n$ and $Z$ straight into equation (1). For hydrogen ($Z=1$) the $n=3$ orbit has $r_3 = 0.529\times9 = 4.76$ Å, $v_3 = 2.19\times10^6/3 = 0.73\times10^6$ m/s and $E_3 = -13.6/9 = -1.51$ eV. One substitution each.

Find the wavelength of a transition. Compute the gap with equation (2), $\Delta E = 13.6\,Z^2(1/n_1^2 - 1/n_2^2)$, then $\lambda = 1240/\Delta E$. Keep $n_1$ as the lower level so the bracket is positive. Absorption is the same arithmetic run upward: a photon of exactly $\Delta E$ lifts the electron from $n_1$ to $n_2$.

Find an ionisation energy. Read off $|E_n|$ from equation (3): the depth of the rung you start on. Watch the two variables the examiner changes to trip you — the starting level $n$ (divides by $n^2$) and the charge $Z$ (multiplies by $Z^2$).

Count the spectral lines and sort them into series. Equation (4) gives the total, $n(n-1)/2$. Then group by landing level: jumps that end on $n_1=1$ are the Lyman series (ultraviolet), on $n_1=2$ the Balmer series (the visible lines), on $n_1=3$ the Paschen series (infrared). The series limit — the shortest-wavelength edge of a series — is the jump from $n=\infty$, i.e. $\Delta E = 13.6/n_1^2$.

Scale to a hydrogen-like ion. Multiply every energy by $Z^2$, divide every radius by $Z$, multiply every speed by $Z$, and divide every wavelength by $Z^2$. He⁺ ($Z=2$) is the stock example: ground energy $-54.4$ eV, first Bohr radius $0.264$ Å, and a Balmer-$\alpha$ line at $164$ nm.

1 · Board level — the nth orbit of hydrogen (CBSE and NEET pattern)

For the hydrogen atom, find the radius, the electron's speed, and the total energy of the second Bohr orbit ($n=2$). Compare each with the ground-state value.

Step 1. Radius, from equation (1): $r_2 = 0.529\,n^2 = 0.529\times4 = \boxed{2.12\ \text{Å}}$ — four times the ground-state $r_1 = 0.529$ Å, because $r\propto n^2$.

Step 2. Speed, from equation (1): $v_2 = 2.19\times10^6/n = 2.19\times10^6/2 = \boxed{1.1\times10^6\ \text{m/s}}$ — half the ground-state $v_1 = 2.19\times10^6$ m/s, because $v\propto 1/n$.

Step 3. Energy, from equation (1): $E_2 = -13.6/n^2 = -13.6/4 = \boxed{-3.4\ \text{eV}}$ — one-quarter as deep as the ground-state $-13.6$ eV, so the electron is less tightly bound.

Step 4. Sense-check: a bigger, slower orbit that is less bound — exactly the trend of the ladder. The electron in $n=2$ is farther out, dawdling, and easier to knock free.

2 · Board level — the Balmer-alpha line and its ionisation partner (ICSE and CBSE pattern)

A hydrogen atom emits light when its electron drops from $n=3$ to $n=2$. Find the energy and wavelength of the emitted photon, and separately state how much energy is needed to ionise a hydrogen atom starting from $n=2$.

Step 1. Energy gap, from equation (2): $\Delta E = 13.6(1/2^2 - 1/3^2) = 13.6(1/4 - 1/9) = 13.6\times0.1389 = \boxed{1.89\ \text{eV}}$. The lower level $n_1=2$ supplies the larger fraction, keeping $\Delta E$ positive.

Step 2. Wavelength, from the photon shortcut: $\lambda = 1240/\Delta E = 1240/1.89 = \boxed{656\ \text{nm}}$ — deep red, the signature Balmer-$\alpha$ line you see in every hydrogen discharge tube.

Step 3. Cross-check with Rydberg: $1/\lambda = R(1/4 - 1/9) = 1.097\times10^7\times0.1389 = 1.524\times10^6$ m⁻¹, so $\lambda = 656$ nm — same answer, second road.

Step 4. Ionisation from $n=2$, from equation (3): $E_{\text{ion}}(2) = |E_2| = 13.6/4 = \boxed{3.4\ \text{eV}}$ — not 13.6 eV, because the electron started halfway up the well, not at the bottom.

3 · Competitive level — count the lines and find the series limits (JEE-Main and NEET pattern)

A sample of atomic hydrogen is excited so that every atom reaches the $n=4$ level. How many distinct spectral lines can the gas emit as the electrons cascade down? Identify the shortest-wavelength line of the resulting spectrum, and quote the series limit of the Balmer series.

Step 1. Line count, from equation (4): $N = n(n-1)/2 = 4\times3/2 = \boxed{6}$ distinct lines — the six ways of pairing the four rungs $\{1,2,3,4\}$.

Step 2. The shortest wavelength is the biggest energy jump, which is $4\to1$ (a Lyman line landing on the ground state): $\Delta E = 13.6(1 - 1/16) = 13.6\times0.9375 = 12.75$ eV, so $\lambda = 1240/12.75 = \boxed{97.3\ \text{nm}}$ — far ultraviolet.

Step 3. The longest wavelength is the smallest jump, $4\to3$ (Paschen): $\Delta E = 13.6(1/9 - 1/16) = 13.6\times0.04861 = 0.661$ eV, so $\lambda = 1240/0.661 = 1876$ nm — infrared. The six lines thus span from $97$ nm (UV) to $1876$ nm (IR).

Step 4. Balmer series limit ($n_2\to\infty$, landing on $n_1=2$): $\Delta E = 13.6/4 = 3.4$ eV, so $\lambda_{\text{limit}} = 1240/3.4 = \boxed{365\ \text{nm}}$ — the near-UV edge where the Balmer lines pile up and the series ends. Timing note: about 90 seconds. Do not compute all six wavelengths — the examiner only wants the count ($n(n-1)/2$) plus the two extremes, which are always the largest jump (to $n=1$) and the smallest jump ($n\to n-1$). Find those two and you have bracketed the whole spectrum.

4 · Competitive level — scaling to a helium ion (JEE-Advanced pattern)

The singly-ionised helium atom He⁺ ($Z=2$) is a one-electron atom like hydrogen. Find the energy and radius of its ground state, its ionisation energy from the ground state, and the wavelength of its Balmer-$\alpha$ ($3\to2$) line. Compare each with hydrogen.

Step 1. Ground-state energy, from equation (1) with $Z=2$: $E_1 = -13.6\,Z^2/n^2 = -13.6\times4/1 = \boxed{-54.4\ \text{eV}}$ — four times deeper than hydrogen's $-13.6$ eV, because $E\propto Z^2$.

Step 2. Ground-state radius: $r_1 = 0.529\,n^2/Z = 0.529/2 = \boxed{0.264\ \text{Å}}$ — half of hydrogen's $0.529$ Å, because $r\propto 1/Z$; the doubled nuclear charge pulls the electron in tighter.

Step 3. Ionisation energy from the ground state, from equation (3): $E_{\text{ion}}(1) = |E_1| = \boxed{54.4\ \text{eV}}$ — you must supply the full well depth to free the electron.

Step 4. Balmer-$\alpha$ wavelength, from equation (2) with $Z=2$: $\Delta E = 13.6\times4\times(1/4 - 1/9) = 54.4\times0.1389 = 7.56$ eV, so $\lambda = 1240/7.56 = \boxed{164\ \text{nm}}$ — hydrogen's $656$ nm red line, divided by $Z^2 = 4$, now a far-UV line. Every He⁺ line is a hydrogen line shifted to one-quarter the wavelength. Timing note: about 60 seconds if you scale rather than recompute. Never redo the arithmetic from constants — take the hydrogen answer and apply the four scalings ($E\times Z^2$, $r/Z$, $v\times Z$, $\lambda/Z^2$). For $Z=2$ that is just "energies $\times4$, radii $\div2$, wavelengths $\div4$", and the whole problem is mental arithmetic.

Dropping the minus sign. Bohr energies are negative: $E_n = -13.6\,Z^2/n^2$ eV. The ground state $n=1$ is the most negative and most bound; higher $n$ is less negative. A positive Bohr energy is always a slip. When you subtract for a transition, write $\Delta E = 13.6\,Z^2(1/n_1^2 - 1/n_2^2)$ with the lower level $n_1$ first, so the result comes out positive automatically.

Quoting $13.6$ eV for the wrong level or atom. $13.6$ eV is the ionisation energy of hydrogen from its ground state only. From $n=2$ it is $3.4$ eV; for He⁺ from its ground state it is $54.4$ eV. Ionisation from level $n$ is $|E_n| = 13.6\,Z^2/n^2$ — check the level and the charge every time.

Miscounting spectral lines. A gas reaching level $n$ emits $n(n-1)/2$ lines, not $n-1$. The value $n-1$ is only the number of lines in a single series (those ending on one particular rung). Excite to $n=4$: six lines, not three.

Forgetting the $Z^2$ (and $1/Z$) scalings. For a hydrogen-like ion, energies scale as $Z^2$, radii as $1/Z$, speeds as $Z$ and wavelengths as $1/Z^2$. The commonest competitive-exam blunder is scaling the wavelength by $Z$ instead of $Z^2$ — He⁺'s Balmer-$\alpha$ is at $164$ nm ($656/4$), not $328$ nm.

Unit slips. Radii come out in ångström ($1$ Å $=10^{-10}$ m), energies in electronvolts, wavelengths in nanometres via $\lambda = 1240/\Delta E$. Convert $\Delta E$ to joules ($\times1.6\times10^{-19}$) only if the question explicitly wants $f = \Delta E/h$ in hertz or an SI wavelength. Carry four significant figures and present two or three.

Board scope. The Bohr model, hydrogen energy levels, the line spectrum and these numerical families are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, and the energy-level and line-spectrum reasoning is on AP Physics 2 (Modern Physics) and A-Level 9702. IB Physics does not carry the quantitative Bohr $r_n$/$v_n$/$E_n$ treatment in its core — energy levels and photon transitions appear (HL more than SL), but the orbit-radius and orbit-speed numericals do not — so this gym is not IB revision. IGCSE 0625 has no quantitative atomic physics at all, so do not revise this lane for an IGCSE paper.

Texts. NCERT Physics Class 12, Ch. 12 ("Atoms"), the Bohr-model derivation and the spectral-series exercises · H. C. Verma, Concepts of Physics Vol. 2, "Bohr's Model and Physics of the Atom," worked problems · D. C. Pandey / Irodov atomic-physics problem sets (for the $Z$-scaling and line-count variants) · Cambridge A-Level 9702, atomic-spectra and energy-level problems.

Provenance. Standard exam-archive problems on Bohr's one-electron atom. Recompute: master ladder $r_n = 0.529\,n^2/Z$ Å, $v_n = 2.19\times10^6\,Z/n$ m/s, $E_n = -13.6\,Z^2/n^2$ eV; $n=2$ hydrogen $\Rightarrow r_2 = 2.12$ Å, $v_2 = 1.1\times10^6$ m/s, $E_2 = -3.4$ eV; $n=3 \Rightarrow r_3 = 4.76$ Å, $v_3 = 0.73\times10^6$ m/s, $E_3 = -1.51$ eV; Balmer-$\alpha$ ($3\to2$) $\Rightarrow \Delta E = 1.89$ eV, $\lambda = 656$ nm (Rydberg cross-check $1/\lambda = 1.524\times10^6$ m⁻¹); Hβ ($4\to2$) $\lambda = 486$ nm; Lyman-$\alpha$ ($2\to1$) $\Rightarrow 10.2$ eV, $122$ nm; ionisation from $n=2 \Rightarrow 3.4$ eV; excite to $n=4 \Rightarrow 6$ lines, shortest $4\to1$ at $97.3$ nm, longest $4\to3$ at $1876$ nm; series limits Lyman $91.2$ nm, Balmer $365$ nm, Paschen $821$ nm; lines from $n=5 \Rightarrow 10$; He⁺ ($Z=2$) $\Rightarrow E_1 = -54.4$ eV, $r_1 = 0.264$ Å, ionisation $54.4$ eV, Balmer-$\alpha$ $164$ nm.

Prerequisites. Energy Levels and the Hydrogen Spectrum — the concept article that builds the $E_n = -13.6/n^2$ ladder, the Rydberg formula and the Lyman/Balmer/Paschen series this gym drills. Leads to. the university seam The Bohr Model of Hydrogen (`aq-optics-bohr-model-hydrogen`), which re-derives the levels and prepares the ground for the full quantum theory.

Go deeper (UG). The rigorous quantum-mechanical account — the hydrogen atom as a solution of the Schrödinger equation, the true orbital shapes, the angular-momentum and spin quantum numbers, and the fine-structure corrections that Sommerfeld first glimpsed — is developed in The Bohr Model of Hydrogen and its successors. That machinery is beyond this article's algebra-and-trig level; here we quote its result, which for a one-electron atom is exactly Bohr's ladder. Enrichment only; nothing in this gym depends on it.