Structure of Atom • Topic 1 of 3

Sub-atomic Particles & Atomic Models

For a long time the atom was believed to be the smallest, indivisible unit of matter — Dalton's word atomos literally means "uncuttable". Experiments with electricity in the late nineteenth century shattered that picture by revealing that atoms themselves are built from still smaller sub-atomic particles: the electron, the proton and the neutron.

The electron was discovered through cathode ray experiments in a discharge tube. J. J. Thomson showed that these rays travel from cathode to anode, are deflected by both electric and magnetic fields towards the positive plate, and behave identically whatever gas or electrode is used — so a negatively charged particle is common to all matter. By balancing the electric and magnetic deflections he measured the charge-to-mass ratio $\frac{e}{m_e}=1.758\times10^{11}\,\text{C kg}^{-1}$. R. A. Millikan's oil-drop experiment later fixed the electron's charge as $e=1.602\times10^{-19}\,\text{C}$, giving the electron mass $m_e=9.11\times10^{-31}\,\text{kg}$.

Positively charged canal rays (anode rays) revealed the proton. Unlike cathode rays, their charge-to-mass ratio depends on the gas, and it is largest for hydrogen — so the hydrogen positive ion (the proton) is the fundamental unit of positive charge, with charge $+1.602\times10^{-19}\,\text{C}$ and mass $1.672\times10^{-27}\,\text{kg}$ (about $1837$ times the electron). The electrically neutral neutron, of nearly the same mass, was discovered by James Chadwick in 1932 when he bombarded beryllium with $\alpha$-particles.

  • Atomic number $Z$ = number of protons = number of electrons in a neutral atom; it fixes the element's identity.
  • Mass number $A$ = protons + neutrons (nucleons). Number of neutrons $= A-Z$.
  • Isotopes: same $Z$, different $A$ (e.g. $^{1}_{1}\text{H}$, $^{2}_{1}\text{H}$, $^{3}_{1}\text{H}$). Isobars: same $A$, different $Z$ (e.g. $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$).

Thomson's model (the "plum-pudding" model) pictured the atom as a uniform sphere of positive charge with electrons embedded in it like seeds in a watermelon. It explained electrical neutrality but failed completely when Geiger and Marsden fired $\alpha$-particles at thin gold foil. Most particles passed straight through, a few were deflected at large angles, and roughly $1$ in $20000$ bounced almost straight back. From this Rutherford concluded that the atom is mostly empty space, with a tiny, dense, positively charged nucleus (radius $\sim10^{-15}\,\text{m}$ against an atomic radius $\sim10^{-10}\,\text{m}$) around which electrons orbit. But classical physics says an accelerating (orbiting) electron must radiate energy and spiral into the nucleus — Rutherford's atom should collapse in $\sim10^{-8}\,\text{s}$, and it could not explain line spectra.

Bohr's model rescued the hydrogen atom with three postulates: the electron moves only in certain stationary orbits without radiating; angular momentum is quantised, $mvr=\frac{nh}{2\pi}$ (with $n=1,2,3,\dots$); and energy is absorbed or emitted only when the electron jumps between orbits, $\Delta E=E_{\text{final}}-E_{\text{initial}}=h\nu$. For hydrogen the allowed energies and radii come out as $E_n=-\frac{13.6}{n^2}\,\text{eV}$ and $r_n=0.529\,n^2\,\text{angstrom}$, so $r_n\propto n^2$. The negative sign means the electron is bound; $13.6\,\text{eV}$ is the ionisation energy of hydrogen.

Bohr's model beautifully reproduces the hydrogen spectrum. When excited electrons fall back, they emit photons of fixed wavelength, grouped into series. The Rydberg equation gives the wavenumber of every line: $\bar{\nu}=\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, with $R_H=1.097\times10^{7}\,\text{m}^{-1}$ and $n_2>n_1$. The Lyman series ($n_1=1$, ultraviolet), Balmer series ($n_1=2$, visible) and Paschen, Brackett, Pfund series ($n_1=3,4,5$, infrared) all follow from one formula. Its limitations: it works only for one-electron species (H, $\text{He}^+$, $\text{Li}^{2+}$), cannot explain fine spectral structure or the splitting of lines in magnetic and electric fields (Zeeman and Stark effects), ignores the wave nature of the electron, and violates the uncertainty principle by assigning a definite orbit.

Hydrogen energy-level diagram showing Lyman, Balmer and Paschen series transitionsHydrogen Energy Levels & Spectral Seriesn=5 -0.54 eVn=4 -0.85 eVn=3 -1.51 eVn=2 -3.40 eVn=1 -13.6 eVLyman (UV)Balmer (visible)Paschen (IR)Arrows = electron falling to a lower level, emitting a photon
1
Worked Example
An atom of an element has $17$ protons and $18$ neutrons. Write its atomic number, mass number, and the number of electrons in the neutral atom.
Solution
  1. Atomic number $Z=$ number of protons $=17$.
  2. Mass number $A=$ protons $+$ neutrons $=17+18=35$.
  3. A neutral atom has electrons $=$ protons $=17$.

Answer: $Z=17$, $A=35$, electrons $=17$ (the element is chlorine, $^{35}_{17}\text{Cl}$).

2
Worked Example
Calculate the energy of the electron in the $n=2$ orbit of a hydrogen atom, and find how much energy is needed to ionise it from this level.
Solution
  1. $E_n=-\dfrac{13.6}{n^2}\,\text{eV}$, so $E_2=-\dfrac{13.6}{4}=-3.40\,\text{eV}$.
  2. Ionisation means raising the electron to $n=\infty$, where $E_\infty=0$.
  3. Energy required $=E_\infty-E_2=0-(-3.40)=3.40\,\text{eV}$.

Answer: $E_2=-3.40\,\text{eV}$; ionisation energy from $n=2$ is $3.40\,\text{eV}$.

3
Worked Example
Find the radius of the third Bohr orbit of hydrogen, given the first Bohr radius $r_1=0.529\,\text{angstrom}$.
Solution
  1. $r_n=r_1\,n^2$ since $r_n\propto n^2$.
  2. For $n=3$: $r_3=0.529\times3^2=0.529\times9$.
  3. $r_3=4.761\,\text{angstrom}$.

Answer: $r_3\approx4.76\,\text{angstrom}$ ($=4.76\times10^{-10}\,\text{m}$).

4
Worked Example
Calculate the wavelength of the spectral line emitted when an electron in a hydrogen atom jumps from $n_2=3$ to $n_1=2$. Take $R_H=1.097\times10^{7}\,\text{m}^{-1}$.
Solution
  1. $\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)=1.097\times10^{7}\left(\dfrac{1}{4}-\dfrac{1}{9}\right)$.
  2. $\dfrac{1}{4}-\dfrac{1}{9}=\dfrac{9-4}{36}=\dfrac{5}{36}=0.1389$.
  3. $\dfrac{1}{\lambda}=1.097\times10^{7}\times0.1389=1.524\times10^{6}\,\text{m}^{-1}$.
  4. $\lambda=\dfrac{1}{1.524\times10^{6}}=6.56\times10^{-7}\,\text{m}=656\,\text{nm}$.

Answer: $\lambda\approx656\,\text{nm}$ — the red $\text{H}_\alpha$ line of the Balmer series.

5
Worked Example
The pairs $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$ — are they isotopes or isobars? Give the number of neutrons in each.
Solution
  1. Both have the same mass number $A=40$ but different atomic numbers ($Z=18$ and $Z=20$), so they are isobars.
  2. Neutrons in Ar $=A-Z=40-18=22$.
  3. Neutrons in Ca $=40-20=20$.

Answer: They are isobars; Ar has $22$ neutrons and Ca has $20$ neutrons.

6
Worked Example
Calculate the energy (in joules) released when an electron in hydrogen falls from $n=2$ to $n=1$. Use $1\,\text{eV}=1.602\times10^{-19}\,\text{J}$.
Solution
  1. $E_1=-13.6\,\text{eV}$ and $E_2=-\dfrac{13.6}{4}=-3.40\,\text{eV}$.
  2. $\Delta E=E_2-E_1=-3.40-(-13.6)=10.2\,\text{eV}$ (emitted, so released).
  3. Convert: $10.2\times1.602\times10^{-19}=1.63\times10^{-18}\,\text{J}$.

Answer: $1.63\times10^{-18}\,\text{J}$ ($=10.2\,\text{eV}$, a Lyman-series photon).

Key Points

  • The electron ($e/m_e=1.758\times10^{11}\,\text{C kg}^{-1}$, charge $-1.602\times10^{-19}\,\text{C}$) was found in cathode rays; the proton in anode rays; the neutron by Chadwick.
  • Atomic number $Z=$ protons $=$ electrons (neutral atom); mass number $A=$ protons $+$ neutrons; neutrons $=A-Z$.
  • Isotopes share $Z$ but differ in $A$; isobars share $A$ but differ in $Z$.
  • Bohr's hydrogen atom: quantised angular momentum $mvr=\frac{nh}{2\pi}$, $E_n=-\frac{13.6}{n^2}\,\text{eV}$ and $r_n\propto n^2$.
  • Spectral lines obey $\bar{\nu}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$; Bohr fails for multi-electron atoms and ignores the electron's wave nature.
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Q1.The charge-to-mass ratio $e/m$ of anode (canal) rays is largest when the gas in the discharge tube is:
Explanation: Hydrogen gives the lightest positive ion (the proton), so $e/m$ is maximum, identifying the fundamental unit of positive charge.
Q2.In the gold-foil experiment, the fact that a few $\alpha$-particles bounced almost straight back showed that:
Explanation: Large-angle backscattering requires a tiny, dense, positively charged core — the nucleus — most of the atom being empty space.
Q3.The energy of the electron in the ground state of hydrogen is:
Explanation: $E_n=-\frac{13.6}{n^2}$; for $n=1$, $E_1=-13.6\,\text{eV}$. The magnitude is hydrogen's ionisation energy.
Q4.Which series of the hydrogen spectrum lies in the visible region?
Explanation: The Balmer series ($n_1=2$) lies in the visible region; Lyman is UV, while Paschen, Brackett and Pfund are infrared.