Atoms

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book, leaving 9 questions.

The named spectral series have been removed — but Exercise 12.8 still asks for them.

Searching the chapter returns zero hits for "Lyman", "Paschen", "Brackett", "Pfund" and "Rydberg". "Balmer" appears exactly once, and only as a historical remark in section 12.3 noting that in 1885 Johann Balmer found an empirical formula.

Section 12.5 is titled "The Line Spectra of the Hydrogen Atom", but it discusses transitions, emission and absorption lines in general terms without ever naming a series or giving the Rydberg formula.

Yet Exercise 12.8 reads: "A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?" — a question that cannot be answered in the terms it asks for using only this chapter. Our solution supplies the names and shows where each transition falls.

The Franck-Hertz experiment has been removed — zero hits.

What remains is complete. Thomson's model, alpha-particle scattering, the impact parameter, distance of closest approach, the instability of the classical atom, all three Bohr postulates, the energy levels, and de Broglie's standing-wave explanation are all present.

2. Alpha-Particle Scattering and the Nuclear Atom (Textbook 12.2)

Thomson's model pictured the atom as a uniform sphere of positive charge with electrons embedded in it. Since the positive charge was spread out, it could never produce a strong deflecting force, so the model predicts only small-angle scattering.

Geiger and Marsden's experiment fired alpha particles at a thin gold foil. Most passed nearly straight through, but about one in 8000 was deflected through more than $90^{\circ }$, and a few came almost straight back.

Rutherford's remark was that this was "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

What follows from it:

• Most of the atom is empty space, since most alphas pass undeviated.
• All the positive charge and nearly all the mass sit in a tiny central nucleus, since only a concentrated charge can turn an alpha around.
• The nucleus is about $10^{-15}$ m across against an atomic size of $10^{-10}$ m — a factor of $10^5$, so the nucleus occupies about $10^{-15}$ of the atom's volume.

Impact parameter. The perpendicular distance between the alpha's initial path and the nucleus determines how much it is deflected. A small impact parameter means a close approach and a large deflection; head-on collisions give complete backscattering.

Distance of closest approach. For a head-on collision the alpha stops when all its kinetic energy has become electrostatic potential energy:

$K=\frac{1}{4\pi {\epsilon }_0}\frac{(2e)(Ze)}{d}⟹d=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}K}$

This sets an upper bound on the nuclear size.

Why the target must be heavy. Exercise 12.2 asks what would happen with solid hydrogen instead of gold. A proton is lighter than the alpha particle and carries $Z=1$ rather than 79, so it could neither repel the alpha strongly nor resist being pushed aside. The large-angle scattering would vanish and the nucleus would never have been discovered.

3. Why the Classical Atom Fails (Textbook 12.2 to 12.3)

Rutherford's model is right about structure but fatally wrong as classical physics.

The collapse argument. An orbiting electron is continuously accelerating, and classical electromagnetism requires an accelerating charge to radiate. Losing energy, the electron must spiral inwards, reaching the nucleus in about $10^{-8}$ s. Every atom should collapse almost instantly, which plainly does not happen.

The spectrum argument. As the electron spiralled, its orbital frequency would change continuously, so it would emit a continuous spectrum. What is actually observed is a set of sharp discrete lines at fixed wavelengths, characteristic of each element.

These two failures are what Bohr's postulates were constructed to fix.

4. Bohr's Model (Textbook 12.4)

The three postulates, each answering one difficulty:

1. Stationary orbits. The electron can occupy only certain orbits, in which it does not radiate despite accelerating. This simply forbids the collapse.
2. Quantisation of angular momentum. Only orbits satisfying $mvr=\frac{nh}{2\pi }$ are allowed, which selects the discrete set.
3. Frequency condition. Radiation is emitted or absorbed only when the electron jumps between orbits, with $h\nu =E_i-E_f$. This produces sharp lines rather than a continuum.

The results that follow:

$r_n=n^2{r}_1,r_1=0.53\times 10^{-10}\text{m}$

$v_n=\frac{2.18\times 10^6}{n}{\text{m s}}^{-1},E_n=-\frac{13.6}{n^2}\text{eV}$

Reading the scalings. Radius grows as $n^2$, speed falls as $1/n$, and therefore the orbital period grows as $n^3$ — the relation Exercise 12.6 is built on. Energy rises towards zero as $1/n^2$.

Energy relationships in any orbit:

$K=-E,U=2E,U=-2K$

So in the ground state, $E=-13.6$ eV gives $K=+13.6$ eV and $U=-27.2$ eV. This is Exercise 12.4, and it is worth memorising as a pattern rather than rederiving each time.

Why the energy is negative. A negative total energy means the electron is bound. The 13.6 eV needed to raise it to $E=0$ is the ionisation energy of hydrogen.

5. Line Spectra and de Broglie's Explanation (Textbook 12.5 to 12.6)

Transitions produce the lines. When an atom drops from level $n_i$ to $n_f$:

$h\nu =E_{n_i}-E_{n_f}=13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\text{eV}$

Emission lines appear when electrons fall; absorption lines appear as dark gaps in a continuous spectrum when a cool gas absorbs exactly those energies.

The series. Although this edition does not name them, transitions ending on a given level form a family, and Exercise 12.8 requires them:

A useful shortcut for any transition: $\lambda \text{(nm)}=1240/\Delta E\text{(eV)}$.

De Broglie's explanation of the second postulate (12.6). Bohr had to assume angular momentum was quantised. De Broglie showed why: treat the electron as a wave, and a stable orbit is one in which the wave closes on itself, fitting a whole number of wavelengths around the circumference:

$2\pi r=n\lambda =\frac{nh}{mv}⟹mvr=\frac{nh}{2\pi }$

which is exactly Bohr's condition. A non-integer number of wavelengths would interfere destructively with itself and cancel, so only the standing-wave orbits survive. An arbitrary assumption becomes a consequence of wave behaviour.

The correspondence principle. Exercise 12.9 applies Bohr's condition to the Earth orbiting the Sun and gets $n\approx 2.6\times 10^{74}$. Adjacent levels then differ fractionally by about $10^{-74}$, far too finely to detect, so the motion appears perfectly continuous. Quantum results reproduce classical ones at large quantum numbers.

Summary

• Thomson's model spread the positive charge uniformly, so it predicts only small deflections.
• One alpha particle in 8000 was scattered beyond $90^{\circ }$, proving a tiny, massive, highly charged nucleus.
• The nucleus is about $10^{-15}$ m against an atom's $10^{-10}$ m, so the atom is mostly empty space.
• A small impact parameter gives a large deflection; head-on collisions backscatter.
• Distance of closest approach: $d=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}K}$.
• A hydrogen target would show almost no large-angle scattering, since a proton is lighter than the alpha and has $Z=1$.
• The classical Rutherford atom collapses in $10^{-8}$ s and would emit a continuous spectrum — both contradicted by observation.
• Bohr's postulates: non-radiating stationary orbits; $mvr=nh/2\pi$; and $h\nu =E_i-E_f$.
• $r_n=n^2{r}_1$ with $r_1=0.53\times 10^{-10}$ m; $v_n\propto 1/n$; $E_n=-13.6/n^2$ eV.
• The orbital period grows as $n^3$, since radius goes as $n^2$ and speed as $1/n$.
• In any orbit $K=-E$ and $U=2E$, so the ground state has $K=+13.6$ eV and $U=-27.2$ eV.
• Negative total energy means a bound electron; 13.6 eV is the ionisation energy of hydrogen.
• Transition energies: $13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ eV, with $\lambda \text{(nm)}=1240/\Delta E\text{(eV)}$.
• Series by final level: Lyman (UV), Balmer (visible), Paschen, Brackett, Pfund (infrared).
• De Broglie explains the second postulate: a stable orbit fits a whole number of electron wavelengths, $2\pi r=n\lambda$.
• Applying Bohr's rule to the Earth gives $n\approx 2.6\times 10^{74}$, illustrating the correspondence principle.
• The named spectral series, the Rydberg formula and the Franck-Hertz experiment have been removed, though Exercise 12.8 still asks which series are emitted.