Structure of the Atom

By deflecting cathode rays with electric and magnetic fields, Thomson measured their charge-to-mass ratio $q/m$ and showed they are negative particles (electrons) with a very large $q/m$, so far lighter than atoms. This gave the "plum-pudding" model.

Millikan balanced charged oil drops in an electric field and found their charge was always a multiple of $e \approx 1.6\times10^{-19}\text{ C}$, showing charge is quantised and measuring the electron's charge.

Most alphas passing straight through means the atom is mostly empty space; the rare large-angle deflections mean an alpha occasionally hits something tiny, massive and positive, the nucleus. This disproved the plum-pudding model.

$E_2 = -\dfrac{13.6}{2^2} = -\dfrac{13.6}{4} = -3.4\text{ eV}$.

Bohr postulated that electrons in allowed (stationary) states do not radiate, even though classical physics says accelerating charges must. Energy is emitted or absorbed only during transitions between states, which explained atomic stability.

$\lambda = \dfrac{h}{mv} = \dfrac{6.63\times10^{-34}}{(9.1\times10^{-31})(7.3\times10^{6})} = \dfrac{6.63\times10^{-34}}{6.6\times10^{-24}} \approx 1.0\times10^{-10}\text{ m}$ (about the size of an atom).

$E_{\text{photon}} = E_i - E_f = -1.51 - (-3.40) = 1.89\text{ eV}$. This is the red H-alpha line of the Balmer series.

Photon energy equals the gap between levels. Among these, $n=4 \to n=1$ spans the largest gap ($12.75\text{ eV}$), giving the highest-energy, shortest-wavelength photon.

$\dfrac{1}{\lambda} = 1.097\times10^7\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 1.097\times10^7(0.1875) = 2.06\times10^{6}\text{ m}^{-1}$ (giving $\lambda = 486\text{ nm}$).

Held stationary means $qE = mg$, so $q = \dfrac{mg}{E} = \dfrac{3.2\times10^{-14}}{2.0\times10^{5}} = 1.6\times10^{-19}\text{ C}$, the charge of a single electron.

$\lambda = \dfrac{h}{mv}$ is inversely proportional to mass. At the same speed the lighter electron has the longer wavelength.

A stable orbit needs the electron wave to join up smoothly, so the circumference must hold a whole number of wavelengths: $n\lambda = 2\pi r$. Substituting $\lambda = h/mv$ gives Bohr's condition $mvr = n\dfrac{h}{2\pi}$.

$E = 1.89\times1.6\times10^{-19} = 3.02\times10^{-19}\text{ J}$. Then $\lambda = \dfrac{hc}{E} = \dfrac{1.989\times10^{-25}}{3.02\times10^{-19}} = 6.58\times10^{-7}\text{ m} = 658\text{ nm}$ (red).

Ionisation means reaching $E = 0$ (at $n = \infty$). From $n = 2$: $\Delta E = 0 - (-3.4) = 3.4\text{ eV}$. (From the ground state it would be $13.6\text{ eV}$, the trap.)

Balmer (visible) lines are emissions ending at $n = 2$ (from $n = 3, 4, 5, \ldots$). Transitions ending at $n = 1$ (Lyman) are ultraviolet; those ending at $n = 3$ (Paschen) are infrared.

$\dfrac{1}{\lambda} = \dfrac{1}{434\times10^{-9}} = 2.30\times10^{6}$. Then $\dfrac{1}{R} \cdot \dfrac{1}{\lambda} = 0.210 = 0.25 - \dfrac{1}{n_i^2}$, so $\dfrac{1}{n_i^2} = 0.04$, giving $n_i^2 = 25$ and $n_i = 5$.

Kinetic energy $= eV = (1.6\times10^{-19})(100) = 1.6\times10^{-17}\text{ J}$. Momentum $p = \sqrt{2mE} = \sqrt{2(9.1\times10^{-31})(1.6\times10^{-17})} = 5.4\times10^{-24}$. Then $\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{5.4\times10^{-24}} \approx 1.2\times10^{-10}\text{ m}$.

The orbit holds $n = 2$ whole wavelengths, so $r = \dfrac{n\lambda}{2\pi} = \dfrac{2(6.6\times10^{-10})}{2\pi} = \dfrac{1.32\times10^{-9}}{6.28} = 2.1\times10^{-10}\text{ m}$. (Leaving out the $2\pi$ gives the trap $1.3\times10^{-9}$.)

Charge: $q = \dfrac{mg}{E} = \dfrac{4.8\times10^{-14}}{1.0\times10^{5}} = 4.8\times10^{-19}\text{ C}$. Number of electrons: $\dfrac{q}{e} = \dfrac{4.8\times10^{-19}}{1.6\times10^{-19}} = 3$.

First $v = \dfrac{E}{B} = \dfrac{1.2\times10^{4}}{2.0\times10^{-3}} = 6.0\times10^{6}\text{ m/s}$. In the magnetic field $qvB = \dfrac{mv^2}{r}$, so $\dfrac{q}{m} = \dfrac{v}{rB} = \dfrac{6.0\times10^{6}}{(1.7\times10^{-2})(2.0\times10^{-3})} \approx 1.8\times10^{11}\text{ C/kg}$, the electron's charge-to-mass ratio.

Bohr's model works for hydrogen (one electron) but fails for multi-electron atoms, fine structure, relative line intensities, and the Zeeman effect (line splitting in a magnetic field). A full quantum-mechanical treatment was needed.

Schrödinger replaced Bohr's definite orbits with a wavefunction $\psi$ whose square gives the probability density (orbitals, or "electron clouds"). This model handles multi-electron atoms and line intensities that the Bohr model could not.

$\lambda = \dfrac{h}{mv}$. The ball's enormous mass makes $\lambda \sim 10^{-34}\text{ m}$, far too small to give observable diffraction. Wave effects show up only when $\lambda$ is comparable to the aperture, as for electrons ($\sim 10^{-10}\text{ m}$).

A bound-to-bound absorption needs the photon energy to match the gap exactly: $\Delta E = -3.4 - (-13.6) = 10.2\text{ eV}$. A photon with slightly more energy is not absorbed by this transition, which is why absorption spectra show sharp lines rather than broad bands.

$E_4 = -0.85\text{ eV}$, $E_1 = -13.6\text{ eV}$, so $\Delta E = 12.75\text{ eV} = 2.04\times10^{-18}\text{ J}$. Then $\lambda = \dfrac{hc}{\Delta E} = \dfrac{1.989\times10^{-25}}{2.04\times10^{-18}} \approx 9.7\times10^{-8}\text{ m} = 97\text{ nm}$ (an ultraviolet Lyman line).