Structure of Atom
Subatomic particles, Bohr model, quantum numbers & electronic configuration
Subatomic Particles and Early Atomic Models
Subatomic Particles and the Nuclear AtomTopic 1
Although Dalton imagined atoms as indivisible, a series of experiments revealed that atoms contain smaller subatomic particles. The electron was discovered through cathode-ray experiments (J.J. Thomson), who measured its charge-to-mass ratio; Millikan's oil-drop experiment then gave its charge, $-1.6\times10^{-19}\ \text{C}$. The proton (positive) was found in anode rays, and the neutral neutron was discovered later by Chadwick.
These particles define an atom's identity. The atomic number $Z$ is the number of protons (equal to electrons in a neutral atom) and fixes which element it is. The mass number $A$ is the total number of protons and neutrons (nucleons), so the number of neutrons is $A - Z$. Atoms of the same element with different neutron numbers are isotopes (same $Z$, different $A$), while atoms of different elements with the same mass number are isobars.
Thomson's early plum-pudding model pictured electrons embedded in a sphere of positive charge. This was overturned by Rutherford's alpha-scattering experiment, in which most alpha particles passed through a gold foil but a few were deflected sharply. He concluded that the atom is mostly empty space with a tiny, dense, positively charged nucleus at the centre, around which electrons revolve.
Rutherford's nuclear model, however, could not explain atomic stability: classical physics predicts that a revolving (accelerating) electron should radiate energy and spiral into the nucleus. It also could not account for the discrete line spectrum of hydrogen. These two failures set the stage for Bohr's model, and recognising exactly why Rutherford's model failed is a recurring NEET conceptual question.
| Quantity / particle | Detail |
|---|---|
| Atomic number $Z$ | number of protons |
| Mass number $A$ | protons + neutrons; neutrons $= A - Z$ |
| Isotopes | same $Z$, different $A$ |
| Isobars | same $A$, different $Z$ |
An atom has atomic number $17$ and mass number $35$. How many protons, electrons and neutrons does it have (neutral atom)?
Show solution
Protons $= Z = 17$; electrons $= 17$ (neutral); neutrons $= A - Z = 35 - 17 = 18$.
Why did Rutherford's model fail to explain atomic stability?
Show solution
A revolving electron is accelerating and, by classical physics, should continuously radiate energy and spiral into the nucleus. This would make the atom unstable, which contradicts reality.
The atomic number of an element equals the number of:
Isotopes have the same:
Rutherford's experiment used the scattering of:
The number of neutrons in an atom is:
Atoms with the same mass number but different atomic numbers are:
NEET tip: $Z$ identifies the element; neutrons $= A - Z$. Isotopes (same $Z$) vs isobars (same $A$). Rutherford's model failed on stability and on the line spectrum — both fixed by Bohr.
Bohr Model and the Hydrogen SpectrumTopic 2
Bohr's model rescued the nuclear atom for hydrogen with a set of postulates. Electrons revolve only in certain stationary orbits without radiating energy; the allowed orbits are those in which the angular momentum is quantised as $L = \dfrac{nh}{2\pi}$ (where $n = 1, 2, 3, \dots$); and energy is emitted or absorbed only when an electron jumps between orbits, with $\Delta E = h\nu$.
For the hydrogen atom these postulates give quantised orbits and energies. The radius grows as $r_n \propto n^{2}$, while the energy is negative and given by $E_n = -\dfrac{13.6}{n^{2}}\ \text{eV}$. The negative sign means the electron is bound; the ground state ($n = 1$) is the most stable at $-13.6\ \text{eV}$, and the ionisation energy of hydrogen is therefore $13.6\ \text{eV}$.
The model explains the hydrogen line spectrum. When an electron falls from a higher level to a lower one, it emits a photon of a definite energy, giving sharp spectral lines grouped into series: the Lyman series (to $n = 1$, ultraviolet), Balmer (to $n = 2$, visible), and Paschen, Brackett, Pfund (to $n = 3, 4, 5$, infrared). The wavelengths follow the Rydberg relation, $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$.
Bohr's model works beautifully for hydrogen and other one-electron species (such as $\text{He}^+$, $\text{Li}^{2+}$) but fails for multi-electron atoms, cannot explain the splitting of spectral lines in magnetic fields (Zeeman effect), and violates the later Heisenberg uncertainty principle by assigning electrons definite orbits. These limitations led to the quantum-mechanical model, and knowing both the successes and the failures of Bohr's model is heavily examined in NEET.
| Quantity (hydrogen) | Relation |
|---|---|
| Angular momentum | $L = \dfrac{nh}{2\pi}$ |
| Energy level | $E_n = -\dfrac{13.6}{n^{2}}\ \text{eV}$ |
| Orbit radius | $r_n \propto n^{2}$ |
| Spectral lines | $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$ |
Find the energy of the electron in the third orbit ($n = 3$) of hydrogen.
Show solution
$E_3 = -\dfrac{13.6}{3^{2}} = -\dfrac{13.6}{9} \approx -1.51\ \text{eV}$.
To which spectral series does a transition ending at $n = 2$ in hydrogen belong, and in which region does it lie?
Show solution
It belongs to the Balmer series, which lies in the visible region of the spectrum.
In Bohr's model, angular momentum is quantised as:
The energy of the $n$-th level of hydrogen is:
The Balmer series of hydrogen lies in the:
Bohr's model works well for:
The ionisation energy of ground-state hydrogen is:
NEET tip: Memorise $E_n = -13.6/n^2$ eV, $r_n \propto n^2$, $L = nh/2\pi$. Lyman (UV), Balmer (visible), Paschen (IR). Bohr works only for one-electron species.
Quantum Mechanical Model of the Atom
Dual Nature, Uncertainty and Quantum NumbersTopic 3
The quantum-mechanical model began with two revolutionary ideas. de Broglie proposed that matter has a wave nature, with wavelength $\lambda = \dfrac{h}{mv} = \dfrac{h}{p}$ — significant only for very light particles like electrons. Heisenberg's uncertainty principle states that the position and momentum of an electron cannot both be known precisely at the same time, $\Delta x\cdot\Delta p \geq \dfrac{h}{4\pi}$. Together these dismantled Bohr's fixed orbits, replacing them with regions of probability.
In the modern model, an electron is described by a wave function whose square gives the probability of finding it in a region of space — an orbital. An orbital is not a path but a three-dimensional region where the probability of finding an electron is high. Each orbital and each electron is specified by a set of four quantum numbers.
The principal quantum number $n$ ($1, 2, 3, \dots$) gives the shell and the main energy and size. The azimuthal quantum number $l$ ($0$ to $n-1$) gives the subshell and orbital shape ($l = 0, 1, 2, 3$ are $s, p, d, f$). The magnetic quantum number $m_l$ ($-l$ to $+l$) gives the orbital's orientation, so there are $2l+1$ orbitals in a subshell. The spin quantum number $m_s$ ($+\tfrac{1}{2}$ or $-\tfrac{1}{2}$) gives the electron's spin direction.
The shapes follow from $l$: $s$-orbitals are spherical, $p$-orbitals are dumb-bell shaped (three of them, along $x$, $y$, $z$), and $d$-orbitals are more complex (five of them). A useful NEET fact is the number of nodes: an orbital has $(n-1)$ total nodes, of which $l$ are angular and $(n-l-1)$ are radial. Quantum numbers, their allowed values, and the count of orbitals/electrons per shell are dependable sources of NEET marks.
| Quantum number | Gives / range |
|---|---|
| Principal $n$ | shell, size; $1, 2, 3, \dots$ |
| Azimuthal $l$ | shape; $0$ to $n-1$ ($s,p,d,f$) |
| Magnetic $m_l$ | orientation; $-l$ to $+l$ ($2l+1$) |
| Spin $m_s$ | $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$ |
How many orbitals are present in the $p$ subshell, and how many electrons can it hold?
Show solution
For $p$, $l = 1$, so the number of orbitals is $2l+1 = 3$. Each orbital holds 2 electrons, so the $p$ subshell holds $3\times2 = 6$ electrons.
What is the maximum number of electrons in a shell with principal quantum number $n$?
Show solution
The maximum number of electrons in a shell is $2n^{2}$ — for example, $n = 3$ holds $2\times9 = 18$ electrons.
The de Broglie wavelength is:
The azimuthal quantum number $l$ for a $d$ subshell is:
The number of orbitals in a subshell is:
The maximum number of electrons in a shell is:
The shape of an $s$-orbital is:
NEET tip: Know the four quantum numbers and their ranges. Orbitals per subshell $= 2l+1$; electrons per shell $= 2n^2$; total nodes $= n-1$ (angular $= l$, radial $= n-l-1$).
Electronic Configuration and Filling RulesTopic 4
The arrangement of electrons in an atom's orbitals — its electronic configuration — is governed by three rules. The Aufbau principle states that electrons fill orbitals in order of increasing energy, lowest first. The energy order follows the $(n+l)$ rule: orbitals fill in order of increasing $(n+l)$, and for equal $(n+l)$ the one with lower $n$ fills first. This gives the familiar sequence $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, \dots$, where $4s$ fills before $3d$.
The Pauli exclusion principle states that no two electrons in an atom can have all four quantum numbers identical; in practice, an orbital holds at most two electrons, and they must have opposite spins. Hund's rule of maximum multiplicity states that electrons fill degenerate (equal-energy) orbitals singly first, with parallel spins, before any pairing begins — minimising electron-electron repulsion.
These rules let you write the configuration of any atom, for example $\text{O}\ (Z=8): 1s^2\,2s^2\,2p^4$ and $\text{Na}\ (Z=11): 1s^2\,2s^2\,2p^6\,3s^1$. Configurations are often abbreviated using the nearest preceding noble gas, e.g. $\text{Na} = [\text{Ne}]\,3s^1$. This is a high-frequency NEET skill, and many periodic-table trends follow directly from it.
An important exception arises from the extra stability of half-filled and fully filled subshells. Chromium is $[\text{Ar}]\,3d^5\,4s^1$ (not $3d^4\,4s^2$) and copper is $[\text{Ar}]\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$), because exactly half-filled ($d^5$) and fully filled ($d^{10}$) subshells are unusually stable due to symmetry and exchange energy. These two anomalies are perennial NEET favourites and should be memorised.
| Rule | Statement |
|---|---|
| Aufbau ($n+l$ rule) | fill lowest energy first ($4s$ before $3d$) |
| Pauli exclusion | max 2 electrons/orbital, opposite spins |
| Hund's rule | singly fill degenerate orbitals first |
| Stability | half-filled / fully filled subshells |
Write the electronic configuration of sodium ($Z = 11$).
Show solution
$\text{Na} = 1s^2\,2s^2\,2p^6\,3s^1$, or in shorthand $[\text{Ne}]\,3s^1$.
Why is the configuration of chromium ($Z = 24$) $[\text{Ar}]\,3d^5\,4s^1$ rather than $[\text{Ar}]\,3d^4\,4s^2$?
Show solution
A half-filled $3d^5$ subshell has extra stability (from symmetry and exchange energy), so one $4s$ electron shifts to $3d$ to give the more stable $3d^5\,4s^1$ arrangement.
According to the Aufbau principle, the orbital filled first is:
The Pauli exclusion principle limits an orbital to:
Hund's rule says degenerate orbitals are first filled:
The configuration of copper ($Z = 29$) is:
Extra stability is associated with subshells that are:
NEET tip: Fill by the $(n+l)$ rule ($4s$ before $3d$), obey Pauli (2 per orbital) and Hund (singly first). Remember the Cr ($3d^5 4s^1$) and Cu ($3d^{10} 4s^1$) anomalies.
Quick Revision — Structure of Atom
- Particles: $Z$ = protons (defines element); neutrons $= A - Z$. Isotopes (same $Z$), isobars (same $A$).
- Rutherford: tiny dense nucleus; failed on stability and the line spectrum.
- Bohr (H only): $L = nh/2\pi$, $E_n = -13.6/n^2\ \text{eV}$, $r_n \propto n^2$. Series: Lyman (UV), Balmer (visible), Paschen (IR).
- Quantum numbers: $n$ (shell), $l$ (shape, $0$–$(n-1)$), $m_l$ (orientation, $2l+1$ orbitals), $m_s$ ($\pm\tfrac12$). Electrons per shell $= 2n^2$.
- Filling: Aufbau ($n+l$ rule, $4s$ before $3d$), Pauli (2/orbital), Hund (singly first). Anomalies: $\text{Cr} = 3d^5 4s^1$, $\text{Cu} = 3d^{10} 4s^1$.
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