Structure of Atom • Topic 2 of 3

Quantum Mechanical Model

Bohr's neat orbits could not survive two revolutionary ideas of the 1920s: that matter has a wave nature, and that the very act of locating a tiny particle disturbs it. Together these led to the quantum mechanical model, in which an electron is not a planet on a track but a smeared-out probability cloud described by a wavefunction.

In 1924 Louis de Broglie proposed that every moving particle has an associated wavelength $\lambda=\frac{h}{mv}=\frac{h}{p}$, where $h=6.626\times10^{-34}\,\text{J s}$ is Planck's constant and $p=mv$ is momentum. For everyday objects the mass is so large that $\lambda$ is utterly negligible, which is why we never see a cricket ball diffract. For an electron, however, $\lambda$ is comparable to atomic spacings — and electron diffraction (Davisson and Germer) confirmed it. De Broglie's relation also explains Bohr's quantisation: a stable orbit must contain a whole number of electron wavelengths, $2\pi r=n\lambda$, which rearranges to $mvr=\frac{nh}{2\pi}$.

In 1927 Werner Heisenberg showed there is a fundamental limit on how precisely we can simultaneously know a particle's position and momentum: $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$. The more sharply we pin down position ($\Delta x$ small), the more uncertain the momentum ($\Delta p$ large), and vice versa. This is not a measuring-instrument flaw but a property of nature. It demolishes the idea of a definite electron orbit — if position were exactly known, momentum would be completely undefined. So instead of a path we speak of the probability of finding the electron in a region, described by an orbital: a three-dimensional region around the nucleus where the probability of finding an electron is high (conventionally $90\%$).

Erwin Schrödinger's wave equation, solved for the hydrogen atom, gives wavefunctions $\psi$ labelled by three quantum numbers; a fourth describes the electron's own spin:

  • Principal quantum number $n$ ($=1,2,3,\dots$): fixes the main energy level / shell and the size of the orbital. The shell holds at most $2n^2$ electrons.
  • Azimuthal (angular momentum) quantum number $l$ ($=0$ to $n-1$): fixes the sub-shell and the shape of the orbital. $l=0,1,2,3$ are the $s,p,d,f$ sub-shells.
  • Magnetic quantum number $m_l$ ($=-l$ to $+l$, including $0$): fixes the orientation of the orbital in space, giving $2l+1$ orbitals per sub-shell.
  • Spin quantum number $m_s$ ($=+\tfrac{1}{2}$ or $-\tfrac{1}{2}$): the two possible spin states of an electron.

The shapes follow from $l$. An $s$ orbital ($l=0$) is spherically symmetric. A $p$ orbital ($l=1$) is dumb-bell shaped with two lobes along an axis, and there are three of them ($p_x,p_y,p_z$). The five $d$ orbitals ($l=2$) have more complex four-lobe (and one doughnut-plus-lobes) shapes. Each orbital has a node — a surface where the probability of finding the electron is zero. There are two kinds: radial nodes (spherical, counted by $n-l-1$) and angular nodes (planar/conical, counted by $l$). The total number of nodes is $n-1$. For example a $2s$ orbital has $2-0-1=1$ radial node and $0$ angular nodes; a $2p$ orbital has $0$ radial and $1$ angular node (the nodal plane through the nucleus separating its two lobes).

This probabilistic, wave-based picture replaces fixed Bohr orbits while keeping the quantised energy levels that explained the hydrogen spectrum — but now it works for many-electron atoms too, because the quantum numbers give a complete, consistent address for every electron.

Shapes of s and p atomic orbitalsShapes of s and p Orbitalss (l=0)p_zp_xp_y
The four quantum numbers and what they describe
SymbolNameAllowed valuesDescribes
nPrincipal1, 2, 3, …Shell, size & energy
lAzimuthal0 to (n−1)Sub-shell & shape (s,p,d,f)
mlMagnetic−l to +lOrientation (2l+1 orbitals)
msSpin+½ or −½Electron spin direction
1
Worked Example
Calculate the de Broglie wavelength of an electron ($m=9.11\times10^{-31}\,\text{kg}$) moving with a velocity of $2.0\times10^{6}\,\text{m s}^{-1}$. Take $h=6.626\times10^{-34}\,\text{J s}$.
Solution
  1. $\lambda=\dfrac{h}{mv}$.
  2. $mv=9.11\times10^{-31}\times2.0\times10^{6}=1.822\times10^{-24}\,\text{kg m s}^{-1}$.
  3. $\lambda=\dfrac{6.626\times10^{-34}}{1.822\times10^{-24}}=3.64\times10^{-10}\,\text{m}$.

Answer: $\lambda\approx3.64\times10^{-10}\,\text{m}=3.64\,\text{angstrom}$.

2
Worked Example
A cricket ball of mass $0.15\,\text{kg}$ moves at $40\,\text{m s}^{-1}$. Find its de Broglie wavelength and comment.
Solution
  1. $\lambda=\dfrac{h}{mv}=\dfrac{6.626\times10^{-34}}{0.15\times40}$.
  2. $mv=6.0\,\text{kg m s}^{-1}$.
  3. $\lambda=\dfrac{6.626\times10^{-34}}{6.0}=1.10\times10^{-34}\,\text{m}$.

Answer: $\lambda\approx1.1\times10^{-34}\,\text{m}$ — far too small to detect, so wave effects are negligible for macroscopic objects.

3
Worked Example
An electron's position is known to an accuracy of $\Delta x=1.0\times10^{-10}\,\text{m}$. Find the minimum uncertainty in its momentum. Use $h=6.626\times10^{-34}\,\text{J s}$.
Solution
  1. $\Delta x\cdot\Delta p\ge\dfrac{h}{4\pi}$, so $\Delta p\ge\dfrac{h}{4\pi\,\Delta x}$.
  2. $\dfrac{h}{4\pi}=\dfrac{6.626\times10^{-34}}{12.566}=5.27\times10^{-35}\,\text{J s}$.
  3. $\Delta p\ge\dfrac{5.27\times10^{-35}}{1.0\times10^{-10}}=5.27\times10^{-25}\,\text{kg m s}^{-1}$.

Answer: $\Delta p\ge5.27\times10^{-25}\,\text{kg m s}^{-1}$.

4
Worked Example
For the principal quantum number $n=3$, list the allowed values of $l$ and the sub-shells, and state how many orbitals there are in total.
Solution
  1. $l$ runs from $0$ to $n-1$, so $l=0,1,2$, i.e. $3s,3p,3d$.
  2. Orbitals per sub-shell $=2l+1$: $3s$ has $1$, $3p$ has $3$, $3d$ has $5$.
  3. Total $=1+3+5=9$ orbitals (equal to $n^2=9$).

Answer: $l=0,1,2$ ($3s,3p,3d$); $9$ orbitals in all, holding up to $2n^2=18$ electrons.

5
Worked Example
How many radial nodes, angular nodes and total nodes does a $3p$ orbital have?
Solution
  1. For $3p$: $n=3$, $l=1$.
  2. Radial nodes $=n-l-1=3-1-1=1$.
  3. Angular nodes $=l=1$.
  4. Total nodes $=n-1=2$ (check: $1+1=2$).

Answer: $1$ radial node, $1$ angular node, $2$ total nodes.

6
Worked Example
Using $2\pi r=n\lambda$ and de Broglie's relation, show that the angular momentum of a Bohr electron is quantised.
Solution
  1. A standing-wave orbit needs a whole number of wavelengths: $2\pi r=n\lambda$.
  2. Substitute de Broglie $\lambda=\dfrac{h}{mv}$: $2\pi r=\dfrac{nh}{mv}$.
  3. Rearrange: $mvr=\dfrac{nh}{2\pi}$.

Answer: $mvr=\dfrac{nh}{2\pi}$ — exactly Bohr's quantisation condition, now justified by wave nature.

Key Points

  • Matter has a wave nature: $\lambda=\frac{h}{mv}$; significant for electrons, negligible for macroscopic bodies.
  • Heisenberg's uncertainty principle: $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$ — position and momentum cannot both be exact, so fixed orbits are impossible.
  • An orbital is a region of high probability (≈90%) of finding an electron; it is described by a wavefunction $\psi$.
  • Four quantum numbers: $n$ (size/energy), $l$ (shape, 0 to $n-1$), $m_l$ (orientation, $-l$ to $+l$), $m_s$ ($\pm\tfrac{1}{2}$).
  • $s$ orbitals are spherical, $p$ orbitals dumb-bell shaped; radial nodes $=n-l-1$, angular nodes $=l$, total nodes $=n-1$.
Tap an option to check your answer0 / 4
Q1.The de Broglie wavelength of a particle is given by:
Explanation: $\lambda=\frac{h}{mv}=\frac{h}{p}$: wavelength is inversely proportional to momentum.
Q2.Heisenberg's uncertainty principle states that the product $\Delta x\cdot\Delta p$ is at least:
Explanation: $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$; greater precision in one variable forces greater uncertainty in the other.
Q3.For $n=2$, the allowed values of the azimuthal quantum number $l$ are:
Explanation: $l$ runs from $0$ to $n-1$; for $n=2$ that is $l=0$ ($2s$) and $l=1$ ($2p$).
Q4.The number of angular (nodal-plane) nodes in any $p$ orbital is:
Explanation: Angular nodes $=l$, and for a $p$ orbital $l=1$, so there is exactly one nodal plane through the nucleus.