Chapter 02 · Modern Physics · Full book

Atoms

In the previous chapter we learned to see radiation as grainy and matter as wavy: a beam of light arrives in photons of energy \(h\nu\), and an electron of momentum \(p\) trails a wave of length \(\lambda = h/p\). That double vision was forced on us by experiments—the photoelectric effect on one side, electron diffraction on the other—and it left a promissory note unpaid. We had a new physics of how things travel through empty space. We did not yet have a physics of what they are made of.

This chapter pays the note. It asks the oldest question in natural philosophy—what is matter, at the bottom?—and follows the one experimental thread that finally answered it: fire something small at an atom and watch how it bounces. The answer, when it came, was stranger than anyone had guessed. The atom is almost entirely empty. Nearly all of its mass and all of its positive charge sit in a nucleus ten thousand times smaller than the atom itself, and the electrons occupy the vast remainder. But that picture, taken at face value, is a disaster: a classical atom built this way should destroy itself in about a hundred-billionth of a second. Saving it required a single, audacious quantum rule—and the moment Bohr wrote that rule down, the sharp, coded colours that hydrogen had been emitting for half a century fell out of it exactly, to four-figure accuracy. It was one of the great moments in the history of physics: a guess, a formula, and then nature agreeing to the decimal.

We will earn that formula, not borrow it. As always, the full machinery (the energy levels, the spectral series, the quantisation condition) is built with the calculus and the dynamics a JEE Advanced or Olympiad student needs; the result is then restated in the algebra a NEET or AP Physics 2 student will actually use. And we will be honest, at the end, about where this beautiful model breaks—because the chapter after this one only makes sense once you know that Bohr's atom, for all its triumph, was a scaffold and not the building.

Core intuition: Chapter Roadmap

Early models. Thomson's “plum pudding”: positive charge smeared through the whole atom, electrons stuck in it like raisins. A reasonable guess—and the scattering experiment demolished it.
The \(\alpha\)-scattering experiment. Geiger and Marsden fired \(\alpha\)-particles at gold foil. A few came almost straight back. “As if you fired a shell at tissue paper and it bounced.” That single fact cannot happen with a pudding.
The nuclear atom. A tiny, dense, positive nucleus. We compute the distance of closest approach (how near the \(\alpha\) gets in a head-on hit) and the impact parameter (how the miss-distance fixes the deflection), and read off the size of the nucleus.
The classical catastrophe. An orbiting electron is an accelerating charge; accelerating charges radiate. The classical nuclear atom should spiral into the nucleus in \(\sim10^{-11}\,\)s and glow with every colour as it dies. It does neither. Something is wrong with classical physics, not with the atom.
Bohr's rescue. Three postulates, one of them radical: angular momentum comes only in whole-number steps of \(\hbar\). From that single quantisation we derive the allowed radii, speeds, and energies of hydrogen.
The hydrogen spectrum. The energy levels, turned into emitted wavelengths, reproduce the Lyman, Balmer, and Paschen series and the Rydberg formula—the coded colours, decoded.
Why \(\hbar\)? de Broglie's waves explain the quantisation we had to assume: a stable orbit is one that holds a whole number of electron wavelengths. Then—the limits: why Bohr nails hydrogen and fails for helium.

Perplexing Questions

Perplexing questions

Six puzzles frame this chapter. Each is answered in full at the point where the relevant idea is built—look for the [Resolution: PQ N] callout.

The table you cannot push through. If an atom is almost entirely empty space—a nucleus like a fly in a cathedral, and electrons that are themselves point-like—then matter is mostly nothing. So why can you not push your hand through a tabletop? What is actually stopping you, if there is so little “there” there?
The shell that bounced back. Fire a fast, heavy, positively charged bullet (an \(\alpha\)-particle) at a sheet of gold a few thousand atoms thick. Almost all of them punch straight through, barely deflected. But about one in eight thousand comes back, scattered through more than a right angle. How can a thin foil of light, soft metal turn a heavy bullet around—and what does the rarity of the event tell you about how the atom's charge is arranged?
The atom that refuses to die. An electron circling a nucleus is an accelerating electric charge, and Maxwell's electromagnetism is adamant that accelerating charges radiate energy away. Losing energy, the electron should spiral inward and crash into the nucleus in about a hundred-billionth of a second, emitting a smear of every colour on the way down. Real atoms last forever and emit only a few sharp colours. What is wrong?
The barcode of hydrogen. Heat hydrogen gas and it glows—but not white. Pass its light through a prism and you see only a handful of sharp, isolated lines: a red, a blue-green, a violet, and others crowding toward the ultraviolet, at wavelengths that fit a startlingly simple whole-number formula. Why should the simplest atom emit a code rather than a continuous rainbow—and why those particular wavelengths?
Why whole numbers? Bohr's model works only because he assumed the electron's angular momentum can take the values \(\hbar, 2\hbar, 3\hbar, \dots\) and nothing in between. Quantising angular momentum—of all things—in integer steps is an extraordinary thing to assume. Where could such a rule possibly come from?
One triumph, one failure. The very same model that predicts every line of hydrogen to four-figure accuracy fails completely for the next atom along, helium, with its two electrons. A theory that is exactly right for one case and useless for the next is telling you something. What is the model secretly assuming, and where does that assumption run out?

Look for the [Resolution: PQ N] callout as you read.

From the Indivisible to the Pudding

That matter is made of atoms—discrete, countable units that cannot be cut indefinitely—was an old idea by the time physics could test it. The nineteenth century had made it quantitative: Dalton's chemistry weighed atoms against one another, the kinetic theory of Volume 1 built the gas laws out of them, and by 1897 J. J. Thomson had shown that atoms are not even the smallest things. Firing cathode rays through electric and magnetic fields, he measured the charge-to-mass ratio of their constituent particles and found it enormous—some two thousand times larger than that of the lightest ion. The carriers were a new particle, far lighter than any atom: the electron. The atom had structure.

But the electron is negative, and bulk matter is neutral. Somewhere in the atom there must be a compensating positive charge, and as much of it as the electrons' negative. How is it arranged? Thomson proposed the natural guess, the plum-pudding model: the positive charge is a diffuse sphere filling the whole atom—the “pudding”—with the tiny electrons embedded in it like raisins, distributed so the whole thing is neutral and mechanically stable. It was a serious, testable model. Its central claim is that the atom's positive charge and mass are spread out over the atom's full size, roughly \(10^{-10}\,\)m.

That single claim is what the next experiment was about to destroy. If you want to know whether a charge is spread thin or packed tight, you probe it with another charge and watch how hard it pushes back. The harder and more concentrated the hidden charge, the closer you can get to it and the more violently it can deflect your probe. Rutherford had exactly such a probe to hand.

The Geiger–Marsden Experiment

Radioactive sources emit \(\alpha\)-particles: helium nuclei, charge \(+2e\), mass about \(7300\) times the electron's, moving at a few percent of the speed of light with kinetic energies of several MeV. They are ideal probes—heavy enough not to be deflected by the feather-light electrons, fast enough to drive deep into an atom, and charged enough to feel its positive core.

Under Rutherford's direction, Geiger and Marsden sent a narrow beam of \(\alpha\)-particles at a gold foil only a few thousand atoms thick and counted, with a scintillating screen, how many were scattered through each angle \(\theta\). Gold was chosen because it can be beaten into extraordinarily thin, uniform sheets, so that most \(\alpha\)'s encounter essentially one atom's worth of deflection.

The results split into two utterly different populations.

The overwhelming majority passed straight through, deflected by less than a degree. The atom, to most \(\alpha\)'s, is transparent.
A tiny minority—roughly one in eight thousand—were scattered through angles greater than \(90^\circ\), some bouncing almost straight back.

The first fact is comfortable for either model: an atom is mostly empty and the probe sails through. The second is fatal to the pudding. A diffuse, spread-out positive charge can never deliver the concentrated push needed to reverse a heavy, fast \(\alpha\)-particle. Inside Thomson's pudding the field is gentle everywhere; the largest force the \(\alpha\) could feel is far too weak, and a foil's worth of small random nudges adds up to a fraction of a degree, never a reversal. To turn the \(\alpha\) around you need to bring it very close to a charge that is both large and concentrated into a tiny volume, so that the inverse-square force near it becomes enormous. Rutherford's verdict was famous: “It was as if you fired a fifteen-inch shell at a piece of tissue paper and it came back and hit you.”

Core intuition: [Resolution: PQ 2] The Shell That Bounced Back

The rarity is the message. If the positive charge were spread through the whole atom, no \(\alpha\) could ever be turned around—so a back-scatter rate of even one in eight thousand is impossible for a pudding. The only way to get a violent reversal is to occasionally bring the \(\alpha\) very close to a small, massive, concentrated lump of positive charge. That it happens rarely shows the lump is tiny: almost every \(\alpha\) misses it by a mile (by atomic standards) and barely deflects; only the rare near-head-on hit recoils. A large scattering angle is the signature of a small, dense nucleus, and the small probability of that angle measures how little of the atom the nucleus occupies.

◇ Figure 1

Fig 1. The Geiger–Marsden apparatus. A collimated \(\alpha\) beam strikes a thin gold foil; a movable zinc-sulphide screen counts the scintillations at each scattering angle \(\theta\). The count rate versus \(\theta\) tested the \(\sin^{-4}(\theta/2)\) law.

◇ Figure 2

Fig 2. Two models, two predictions. In Thomson's diffuse atom (left) every \(\alpha\) is barely bent. In Rutherford's nuclear atom (right) most \(\alpha\)'s pass nearly straight, but one approaching the tiny nucleus is flung through a large angle—or straight back. Only the second picture produces the observed back-scattering.

✓ Checkpoint · Early Models and the Scattering Shock

What did the rare large-angle scattering of \(\alpha\)-particles reveal about the atom?
That the positive charge and nearly all the mass are concentrated in a tiny, dense nucleus—not spread out as in the plum-pudding model.
Why does the great majority of the \(\alpha\) beam pass through almost undeflected?
Because the atom is mostly empty space; only the rare \(\alpha\) that approaches the tiny nucleus closely is strongly deflected.
Besides revealing the nucleus, what nuclear property did the scattering data measure for gold?
Its charge: the fit gave \(\approx79e\), matching gold's atomic number \(Z=79\). The atomic number is the nuclear charge.

The Nuclear Atom: Closest Approach and Impact Parameter

Rutherford turned the qualitative shock into a quantitative model. Suppose all the positive charge \(+Ze\) and essentially all the mass sit in a point-like nucleus, and the electrons occupy the surrounding volume. A passing \(\alpha\) of charge \(+2e\) feels only the Coulomb repulsion of the nucleus (the diffuse electrons contribute almost nothing to a fast, heavy probe). Its path is then a hyperbola, exactly the repulsive-Coulomb analogue of the Kepler orbit, with the nucleus at the focus. Two numbers from this picture do all the work.

Distance of closest approach. Consider the cleanest case: a perfectly head-on collision. The \(\alpha\) drives straight at the nucleus, slowing as the repulsive potential energy climbs, until for an instant it stops, all its kinetic energy converted to electrostatic potential energy, and then is flung back. The turning point is the closest it ever gets.

Birth of the formula: Distance of Closest Approach in a Head-On Collision

Let the \(\alpha\)-particle of charge \(q_1 = 2e\) approach a nucleus of charge \(q_2 = Ze\) head-on with kinetic energy \(K = \tfrac12 m v^2\) from far away (where the potential energy is zero). At the turning point its speed is momentarily zero, so all the kinetic energy has become Coulomb potential energy. Energy conservation: \[ K \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0}. \] Solve for the closest-approach distance \(r_0\): \[ \boxed{\,r_0 \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{K}\,} \] Numbers. For \(7.7\,\)MeV \(\alpha\)-particles (the energy Rutherford used) on gold (\(Z=79\)): \(r_0 \approx 3.0\times10^{-14}\,\)m \(= 30\,\)fm. For \(5.0\,\)MeV \(\alpha\)'s, \(r_0 \approx 46\,\)fm. Both are about \(10^4\) times smaller than the atom (\(\sim10^{-10}\,\)m)—the nucleus is a fly in a cathedral. Algebra-level use (NEET / AP Physics 2). The whole result is one line of energy bookkeeping: kinetic energy in \(=\) Coulomb potential energy at the turning point. No calculus is needed; the boxed formula is all you apply.

Misconception alert: Closest Approach Is Not the Nuclear Radius

It is tempting to read \(r_0\) as “the size of the nucleus.” It is not. \(r_0\) is only how close this particular \(\alpha\), with this energy, manages to get before the Coulomb wall turns it back—and it turns back while still outside the nucleus, never touching it. So \(r_0\) is an upper bound on the nuclear radius: the nucleus is at most this big. You pin the radius down only by raising the \(\alpha\) energy until the closest approach shrinks to the point where the deflection stops obeying the pure-Coulomb law—the energy at which the \(\alpha\) finally grazes the nucleus and the short-range nuclear force kicks in. That onset, not \(r_0\) itself, measures the radius.

Impact parameter. A head-on hit is the rare extreme. In general the \(\alpha\) aims a little to one side of the nucleus, missing the centre by a perpendicular distance \(b\) called the impact parameter—the miss-distance the particle would have if there were no force. A large \(b\) means a distant flyby and a gentle deflection; a small \(b\) means a close pass and a sharp one. The exact relation between \(b\) and the scattering angle \(\theta\) is the heart of the model.

Birth of the formula: Impact Parameter and Scattering Angle

The \(\alpha\) moves on a hyperbola in the repulsive Coulomb field of the nucleus. Writing \(K=\tfrac12 mv^2\) for the incoming kinetic energy, the geometry of the hyperbolic orbit (conservation of energy and of angular momentum about the nucleus) gives the scattering angle \(\theta\) in terms of the impact parameter \(b\): \[ \tan\!\frac{\theta}{2} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{2K\,b} \qquad\Longleftrightarrow\qquad \boxed{\,b \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{Ze^2}{K}\,\cot\!\frac{\theta}{2}\,} \] Read the formula physically. As \(b\to 0\) (a near head-on aim), \(\cot(\theta/2)\to0\), so \(\theta\to180^\circ\): the particle comes straight back. As \(b\) grows large, \(\theta\to0\): a distant flyby is barely bent. The smaller the miss-distance, the larger the kick—exactly the trend the experiment showed. Consistency check. Setting \(\theta=180^\circ\) does not give \(b=0\) and the closest approach \(r_0\) of the previous box simultaneously by accident: the head-on particle (\(b=0\)) reaches precisely \(r_0\), while a particle with \(b\gt 0\) has angular momentum and turns back farther out. The two boxed results are the \(b=0\) and \(b\gt 0\) faces of the same hyperbolic mechanics.

The scattering law and nuclear charge. Counting how many \(\alpha\)'s scatter into each angle, Rutherford derived that the number detected at angle \(\theta\) falls off as \[ N(\theta)\;\propto\;\frac{1}{\sin^4(\theta/2)}, \] a spectacularly steep dependence—scattering at \(30^\circ\) is about fourteen times more likely than at \(60^\circ\). Geiger and Marsden confirmed this law over a range of \(10^5\) in count rate, across several foils and \(\alpha\) energies. The fit also returned the nuclear charge: for gold it came out near \(79\,e\), matching gold's atomic number \(Z=79\). The atomic number, until then a mere ordinal in the periodic table, was revealed as a physical quantity—the charge on the nucleus. (We state the \(\sin^{-4}\) law here as the experimental fingerprint; its full derivation from the hyperbolic orbit is developed in the worked examples and problem bank, where it belongs to the JEE Advanced/Olympiad tier.)

Core intuition: [Resolution: PQ 1] The Table You Cannot Push Through

The scattering experiment confirms the atom is mostly empty: a tiny nucleus, point electrons, and a vast void between. So why is matter solid? Not because the nucleus fills space—it manifestly does not—but because of the electrons. When you press your hand on a table, the outer electron clouds of your skin's atoms approach the outer electron clouds of the table's atoms, and like charges repel: the electromagnetic force between those clouds is what your hand feels as solidity. Emptiness and impenetrability are not in conflict; the resistance lives in the fields, not in the matter filling the space. (The full story—why electrons refuse to be squeezed into the same state, the Pauli exclusion principle—is beyond Bohr's atom and beyond this chapter, but the electromagnetic half of the answer is already enough to dissolve the paradox.)

◇ Figure 3

Fig 3. Rutherford scattering geometry. An \(\alpha\)-particle aimed a perpendicular distance \(b\) (the impact parameter) from the nucleus follows a hyperbola and emerges deflected through the scattering angle \(\theta\). A small \(b\) gives a large \(\theta\); a head-on aim (\(b=0\)) is turned straight back (\(\theta=180^\circ\)).

✓ Checkpoint · The Nuclear Atom: Closest Approach and Impact Parameter

A \(4.0\,\mathrm{MeV}\) \(\alpha\)-particle is fired head-on at a copper nucleus (\(Z=29\)). Find its distance of closest approach.
\(r_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}=\dfrac{(8.99\times10^9)(2)(29)(1.6\times10^{-19})^2}{4.0\times10^6\times1.6\times10^{-19}}\approx2.1\times10^{-14}\,\mathrm{m}=21\,\mathrm{fm}\).
As the \(\alpha\)-particle's energy increases, does the distance of closest approach increase or decrease?
Decrease: \(r_0\propto1/K\), so a faster \(\alpha\) pushes deeper before being turned back.
Why is the distance of closest approach only an upper bound on the nuclear radius?
The \(\alpha\) turns back while still outside the nucleus (the Coulomb wall stops it before contact), so the nucleus is at most that large.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Distance of closest approach

A \(5.0\,\mathrm{MeV}\) \(\alpha\)-particle (charge \(+2e\)) is fired head-on at a gold nucleus (\(Z=79\)). Find its distance of closest approach.

Solution. Given: \(K=5.0\,\mathrm{MeV}=8.0\times10^{-13}\,\mathrm{J}\), \(Z=79\), projectile charge \(2e\).
Find: closest approach \(r_0\).
Setup: At the turning point all kinetic energy has become Coulomb potential energy: \(K=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}\).
Solve: \[ r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} =\frac{(8.99\times10^{9})(2)(79)(1.6\times10^{-19})^2}{8.0\times10^{-13}} \approx 4.5\times10^{-14}\,\mathrm{m}. \] Answer: \(\boxed{r_0\approx 4.5\times10^{-14}\,\mathrm{m}=45\,\mathrm{fm}}\)

Example 2 · Energy levels and a transition photon

For hydrogen, find the energies of the \(n=2\) and \(n=4\) levels, and the energy of the photon emitted in a \(4\to2\) transition.

Solution. Given: hydrogen (\(Z=1\)), levels \(n=2,4\).
Find: \(E_2,\,E_4\), and \(h\nu\) for \(4\to2\).
Setup: \(E_n=-13.6/n^2\,\mathrm{eV}\); emitted photon energy \(=E_4-E_2\).
Solve: \[ E_2=-\frac{13.6}{4}=-3.40\,\mathrm{eV},\quad E_4=-\frac{13.6}{16}=-0.85\,\mathrm{eV},\quad h\nu=E_4-E_2=2.55\,\mathrm{eV}. \] Answer: \(\boxed{E_2=-3.40\,\mathrm{eV},\ E_4=-0.85\,\mathrm{eV},\ h\nu=2.55\,\mathrm{eV}}\) (an H\(\beta\) photon, blue-green).

Example 3 · Bohr radii and the role of \(Z\)

Find the radius of the \(n=3\) orbit of hydrogen, and the ground-state radii of He\(^+\) (\(Z=2\)) and Li\(^{2+}\) (\(Z=3\)). Take \(a_0=0.529\,Å\).

Solution. Given: \(a_0=0.529\,Å\), \(r_n=n^2a_0/Z\).
Find: \(r_3(\mathrm{H})\), \(r_1(\mathrm{He^+})\), \(r_1(\mathrm{Li^{2+}})\).
Setup: radius grows as \(n^2\), shrinks as \(1/Z\).
Solve: \[ r_3(\mathrm{H})=9a_0=4.76\,Å,\quad r_1(\mathrm{He^+})=\frac{a_0}{2}=0.265\,Å,\quad r_1(\mathrm{Li^{2+}})=\frac{a_0}{3}=0.176\,Å. \] Answer: \(\boxed{4.76\,Å,\ 0.265\,Å,\ 0.176\,Å}\)

Example 4 · Wavelength of the H\(\alpha\) (Balmer-\(\alpha\)) line

Using the Rydberg formula, find the wavelength of the \(3\to2\) line of hydrogen.

Solution. Given: \(n_f=2,\ n_i=3\), \(R=1.097\times10^7\,\mathrm{m^{-1}}\), \(Z=1\).
Find: \(\lambda\).
Setup: \(\dfrac{1}{\lambda}=R\!\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)\).
Solve: \[ \frac1\lambda=R\!\left(\frac14-\frac19\right)=R\cdot\frac{5}{36} =1.524\times10^{6}\,\mathrm{m^{-1}} \ \Rightarrow\ \lambda=656\,\mathrm{nm}. \] Answer: \(\boxed{\lambda\approx 656\,\mathrm{nm}}\) (the red line; the tabulated value is \(656.3\,\mathrm{nm}\)).

Example 5 · Ionisation energy of H and He\(^+\)

Find the energy needed to remove the electron from the ground state of (a) hydrogen and (b) singly ionised helium He\(^+\).

Solution. Given: ground state \(n=1\); H has \(Z=1\), He\(^+\) has \(Z=2\).
Find: ionisation energies \(=0-E_1=13.6\,Z^2\,\mathrm{eV}\).
Solve: \[ \mathrm{IE(H)}=13.6\,\mathrm{eV},\qquad \mathrm{IE(He^+)}=13.6\times2^2=54.4\,\mathrm{eV}. \] Answer: \(\boxed{13.6\,\mathrm{eV}\ \text{and}\ 54.4\,\mathrm{eV}}\)

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Closest approach:
A \(6.0\,\mathrm{MeV}\) \(\alpha\)-particle is fired head-on at a silver nucleus (\(Z=47\)). Find its distance of closest approach.
\(r_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}=\dfrac{(8.99\times10^9)(2)(47)(1.6\times10^{-19})^2}{6.0\times10^6\times1.6\times10^{-19}}\approx2.3\times10^{-14}\,\mathrm{m}=23\,\mathrm{fm}\).
Energy level of a hydrogen-like ion:
Find the energy of the \(n=2\) level of doubly ionised lithium, \(\mathrm{Li^{2+}}\) (\(Z=3\)).
\(E_2=-13.6\,\dfrac{Z^2}{n^2}=-13.6\,\dfrac{9}{4}=-30.6\,\mathrm{eV}\).
Bohr radius scaling:
Find the radius of the \(n=4\) orbit of hydrogen.
\(r_4=n^2a_0=16\times0.529=8.46\,Å\).
Impact parameter:
A \(7.7\,\mathrm{MeV}\) \(\alpha\)-particle is scattered through \(120^\circ\) by a gold nucleus (\(Z=79\)). Find its impact parameter. (\(\cot60^\circ=1/\sqrt3\).)
\(b=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{K}\cot\dfrac{\theta}{2}=\dfrac{(8.99\times10^9)(79)(1.6\times10^{-19})^2}{7.7\times10^6\times1.6\times10^{-19}}\times\dfrac1{\sqrt3}\approx8.5\times10^{-15}\,\mathrm{m}=8.5\,\mathrm{fm}\).
Virial bookkeeping:
For a hydrogen-like atom, prove \(E=-KE=\tfrac12 PE\), and use it to relate the ionisation energy to the ground-state kinetic energy.
From force balance \(KE=\tfrac12\frac{ke^2 Z}{r}\) and \(PE=-\frac{ke^2Z}{r}=-2\,KE\), so \(E=KE+PE=-KE=\tfrac12 PE\). The ionisation energy is \(0-E=KE_1\), i.e. equal to the ground-state kinetic energy (\(13.6\,\mathrm{eV}\) for H).

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

Derive the Bohr results:
Starting from the Coulomb force balance and the quantisation \(mvr=n\hbar\), derive expressions for \(r_n\), \(v_n\), and \(E_n\) of a hydrogen-like atom.
Force balance \(\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}=\frac{mv^2}{r}\) and \(v=n\hbar/(mr)\) give \(r_n=\frac{4\pi\varepsilon_0 n^2\hbar^2}{mZe^2}=\frac{n^2}{Z}a_0\). Then \(v_n=\frac{Ze^2}{4\pi\varepsilon_0 n\hbar}=\frac{Z}{n}\alpha c\). Energy \(E=\tfrac12 mv^2-\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r}=-\tfrac12\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_n}=-13.6\frac{Z^2}{n^2}\,\mathrm{eV}\).
Read the ladder:
Sketch the hydrogen energy-level diagram and mark the Lyman, Balmer, and Paschen series. Explain in one line why only the Balmer series is (mostly) visible.
Levels at \(E_n=-13.6/n^2\,\mathrm{eV}\) converging on \(0\); Lyman ends on \(n=1\), Balmer on \(n=2\), Paschen on \(n=3\). Only Balmer's transitions (\(n\ge3\to2\)) have photon energies (\(1.9\)–\(3.4\,\mathrm{eV}\)) in the visible band; Lyman is UV (larger gaps to \(n=1\)), Paschen is IR (smaller gaps to \(n=3\)).
The classical catastrophe:
Explain why a classical Rutherford atom is unstable, and state the order of magnitude of its predicted lifetime.
An orbiting electron accelerates; an accelerating charge radiates (Larmor), so it loses energy and spirals inward, the orbital frequency rising continuously (a continuous spectrum). The collapse time is \(\sim10^{-11}\,\mathrm{s}\)—atoms could not exist. Bohr's stationary states (a ground state with no orbit below it) remove the instability.
Closest approach as a bound:
Explain why the distance of closest approach overestimates the nuclear radius, and describe how the true radius is found.
At \(r_0\) the \(\alpha\) is still outside the nucleus and turns back without touching it, so \(r_0\) is only an upper bound. The true radius is found by raising the \(\alpha\) energy (shrinking \(r_0\)) until the scattering deviates from the pure-Coulomb \(\sin^{-4}\) law—the onset of the short-range nuclear force marks the surface.