Atoms and Nuclei • Topic 1 of 3

Atomic Models & Bohr's Theory

The story of the atom begins with Rutherford's nuclear model (1911). In his famous alpha-scattering experiment a thin gold foil was bombarded with $\alpha$-particles. Most passed straight through, a few were deflected through large angles, and about one in 8000 bounced almost straight back. Rutherford concluded that nearly all the mass and the entire positive charge of an atom is concentrated in a tiny central nucleus (radius $\sim 10^{-15}$ m), with electrons orbiting it like planets round the Sun, the rest of the atom being empty space.

The model had two fatal limitations. First, an orbiting electron is an accelerating charge, and classical electromagnetism says such a charge must radiate energy continuously. The electron would therefore spiral into the nucleus in about $10^{-8}$ s — atoms could not be stable. Second, as the electron spiralled in, its frequency of revolution would change continuously, so it should emit a continuous spectrum; but atoms actually emit sharp, discrete line spectra.

In 1913 Niels Bohr rescued the nuclear atom for hydrogen by three bold postulates:

  • Stable orbits: Electrons revolve only in certain allowed circular orbits without radiating energy. These are the stationary states.
  • Quantisation of angular momentum: The angular momentum is an integral multiple of $\frac{h}{2\pi}$, that is $mvr = \frac{nh}{2\pi}$, where $n = 1, 2, 3, \dots$ is the principal quantum number.
  • Frequency condition: Energy is emitted or absorbed only when an electron jumps between orbits, with $h\nu = E_2 - E_1$.

Balancing the Coulomb force against the centripetal force and using the quantisation rule gives the radius of the $n$th orbit: $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$, so that $r_n \propto n^2$. For hydrogen the smallest orbit ($n=1$) is the Bohr radius, $r_1 = 0.53\ \text{\AA}$.

The total energy of the electron in the $n$th level is negative (it is bound) and is given by $E_n = -\frac{13.6}{n^2}\ \text{eV}$. The lowest level $n=1$ ($-13.6$ eV) is the ground state; higher levels are excited states, and $E_\infty = 0$ corresponds to a free electron. The energy needed to remove the electron from the ground state, $13.6$ eV, is the ionisation energy of hydrogen.

When an electron drops from level $n_2$ to a lower level $n_1$, the emitted photon's wavelength follows the Rydberg formula: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, with $R = 1.097 \times 10^7\ \text{m}^{-1}$. Transitions ending on $n_1 = 1$ form the Lyman series (ultraviolet), on $n_1 = 2$ the Balmer series (visible), and on $n_1 = 3$ the Paschen series (infrared). Bohr's model explained the hydrogen spectrum to remarkable accuracy, though it fails for multi-electron atoms.

Hydrogen energy levels and emission series (Lyman, Balmer, Paschen)n=1 E = -13.6 eVn=2 E = -3.4 eVn=3 E = -1.51 eVn=4 E = -0.85 eVn=∞ E = 0Lyman (UV)Balmer (visible)Paschen (IR)
Solved Examples
1
Worked Example
Find the radius of the second Bohr orbit of the hydrogen atom, given that the radius of the first orbit is $0.53\ \text{\AA}$.
Solution
  1. Bohr radii scale as $r_n \propto n^2$, so $r_n = n^2 r_1$.
  2. For $n = 2$: $r_2 = 2^2 \times 0.53 = 4 \times 0.53\ \text{\AA}$.
  3. $r_2 = 2.12\ \text{\AA} = 2.12 \times 10^{-10}\ \text{m}$.

Answer: The second orbit has radius $2.12\ \text{\AA}$.

2
Worked Example
Calculate the energy of the electron in the third excited state ($n = 4$) of hydrogen.
Solution
  1. Use $E_n = -\frac{13.6}{n^2}\ \text{eV}$. The third excited state is $n = 4$.
  2. $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16}\ \text{eV}$.
  3. $E_4 = -0.85\ \text{eV}$.

Answer: $E_4 = -0.85\ \text{eV}$ (negative, since the electron is bound).

3
Worked Example
An electron jumps from $n = 3$ to $n = 2$ in a hydrogen atom. Find the energy and wavelength of the emitted photon.
Solution
  1. $E_2 = -\frac{13.6}{4} = -3.4\ \text{eV}$, $E_3 = -\frac{13.6}{9} = -1.51\ \text{eV}$.
  2. Photon energy $E = E_3 - E_2 = -1.51 - (-3.4) = 1.89\ \text{eV}$.
  3. Wavelength $\lambda = \frac{1240}{E(\text{eV})}\ \text{nm} = \frac{1240}{1.89} \approx 656\ \text{nm}$.

Answer: The photon has energy $1.89\ \text{eV}$ and wavelength $656\ \text{nm}$ — the red $\text{H}_\alpha$ line of the Balmer series.

4
Worked Example
Using the Rydberg formula, find the wavelength of the first line of the Lyman series ($n_2 = 2 \to n_1 = 1$). Take $R = 1.097 \times 10^7\ \text{m}^{-1}$.
Solution
  1. $\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$.
  2. $\frac{1}{\lambda} = \frac{3}{4} \times 1.097 \times 10^7 = 8.23 \times 10^6\ \text{m}^{-1}$.
  3. $\lambda = \frac{1}{8.23 \times 10^6} = 1.215 \times 10^{-7}\ \text{m} = 121.5\ \text{nm}$.

Answer: $\lambda \approx 121.5\ \text{nm}$, lying in the ultraviolet region.

5
Worked Example
What is the ionisation energy of a hydrogen atom in its ground state, and what is meant by it?
Solution
  1. Ionisation energy is the energy needed to take the electron from $n = 1$ to $n = \infty$ (free).
  2. $E_\infty - E_1 = 0 - (-13.6) = 13.6\ \text{eV}$.

Answer: The ionisation energy is $13.6\ \text{eV}$ — the minimum energy required to completely remove the ground-state electron from the atom.

6
Worked Example
An electron in hydrogen absorbs a photon and is excited from the ground state to $n = 3$. Find the wavelength of the absorbed photon.
Solution
  1. $E_1 = -13.6\ \text{eV}$, $E_3 = -\frac{13.6}{9} = -1.51\ \text{eV}$.
  2. Absorbed energy $E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\ \text{eV}$.
  3. $\lambda = \frac{1240}{12.09} \approx 102.6\ \text{nm}$.

Answer: The atom absorbs a photon of wavelength about $102.6\ \text{nm}$ (a Lyman-series line in the UV).

Key Points

  • Rutherford's nuclear model placed all positive charge and most mass in a tiny nucleus, but could not explain atomic stability or line spectra.
  • Bohr's postulates: stable non-radiating orbits, quantised angular momentum $mvr = \frac{nh}{2\pi}$, and emission/absorption only on transitions with $h\nu = E_2 - E_1$.
  • Orbit radius $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$ so $r_n \propto n^2$; for hydrogen $r_1 = 0.53\ \text{\AA}$.
  • Energy levels $E_n = -\frac{13.6}{n^2}\ \text{eV}$; ground state $-13.6$ eV, ionisation energy $13.6$ eV.
  • Rydberg formula $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ gives the Lyman ($n_1=1$, UV), Balmer ($n_1=2$, visible) and Paschen ($n_1=3$, IR) series.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Rutherford's alpha-scattering experiment established the existence of the:
Explanation: The rare large-angle deflections showed charge and mass are concentrated in a tiny central nucleus.
Q2.According to Bohr, the angular momentum of an electron in the $n$th orbit is:
Explanation: Bohr's quantisation rule is $mvr = \frac{nh}{2\pi}$.
Q3.The radius of the $n$th Bohr orbit varies as:
Explanation: $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$, hence $r_n \propto n^2$.
Q4.The Balmer series of hydrogen lies in which region of the spectrum?
Explanation: Balmer transitions end on $n_1 = 2$ and fall in the visible region.
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