MHT-CET Chemistry · Structure of Atom

Quantum Mechanical Model — de Broglie, Heisenberg and Quantum Numbers

The electron is both a wave and a particle: de Broglie gives its wavelength, Heisenberg says you can never pin down its position and momentum together, and four quantum numbers act as the electron's address — naming its shell, subshell, orbital and spin.

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Why this matters

Thirteen PYQs, and the bank tests four reliable patterns: a one-line recall of Heisenberg's principle, a plug-in of de Broglie's formula, a quantum-number-to-orbital label (n=3, l=2 gives 3d), and the (n+l) rule for orbital energy order. Every question is a direct application — no derivations. Learn the four formulas and the l-to-shape mapping and the whole subtopic is arithmetic.

Concept 1 of 5: de Broglie wavelength — wave-particle duality

de Broglie said every moving particle also behaves as a wave, and its wavelength is just Planck's constant divided by its momentum. Heavier or faster particles have shorter wavelengths, which is why the wave nature only shows up for tiny particles like electrons.

Definition

de Broglie's hypothesis links a particle's momentum to a wavelength:

Wavelength $\lambda = \dfrac{h}{mv} = \dfrac{h}{p}$ , where $p = mv$ is the momentum.
Rearranged, the momentum is $p = \dfrac{h}{\lambda}$ — this is the form the bank uses when it gives you $\lambda$ and asks for $p$ .
$\lambda \propto \dfrac{1}{mv}$ : a larger mass or speed means a smaller wavelength, so macroscopic objects have immeasurably tiny wavelengths.
Watch the units: $h = 6.63 \times 10^{-34}\,\text{J s}$ , so with $m$ in kg and $v$ in $\text{m s}^{-1}$ , $\lambda$ comes out in metres.

de Broglie wavelength and momentum

\lambda = \dfrac{h}{mv} = \dfrac{h}{p} \qquad p = \dfrac{h}{\lambda}

$\lambda$ de Broglie wavelength (m)
hPlanck's constant, 6.63e-34 J s
mmass of the particle (kg)
vvelocity of the particle (m/s)
pmomentum, p = mv (kg m/s)

Worked example

Find the de Broglie wavelength of an electron of mass

9.1 \times 10^{-31}\,\text{kg}

moving with a velocity of

2 \times 10^{6}\,\text{m s}^{-1}

. (

h = 6.63 \times 10^{-34}\,\text{J s}

)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q56Moderate

Example 1 · Structure of Atom · Quantum Mechanical Model, de Broglie, Heisenberg and Quantum Numbers

What is de Broglie's wavelength for a particle having mass

6.64 \times 10^{-27}

kg moving with velocity of

3 \times 10^3

m s

^{-1}

? (

h = 6.63 \times 10^{-34}

J s)

Divide by momentum, not by mass alone

The denominator is

mv

(the momentum), not just the mass. Multiply mass by velocity first, then divide

h

by that product.

Convert Ångström to metres

1\,\text{Å} = 10^{-10}\,\text{m}

and

1\,\text{nm} = 10^{-9}\,\text{m}

. A wavelength given in Å or nm must be in metres before you divide, or the power of ten will be wrong.

Drill 1 more on de broglie wavelength — wave-particle duality

Concept 2 of 5: Heisenberg's uncertainty principle

Because an electron is also a wave, you cannot say exactly where it is and exactly how fast it is going at the same instant. The more precisely you fix its position, the fuzzier its momentum becomes, and vice versa — this is a fundamental limit, not a measuring error.

Definition

Heisenberg's uncertainty principle:

It is impossible to determine simultaneously the exact position and the exact momentum of a microscopic particle such as an electron.
The product of the uncertainties has a lower bound: $\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$ .
Equivalently $\Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}$ , since $\Delta p = m\,\Delta v$ .
Small position uncertainty forces a large momentum uncertainty — the two cannot both be zero.

Heisenberg uncertainty relation

\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} \qquad \Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}

$\Delta x$ uncertainty in position
$\Delta p$ uncertainty in momentum
$\Delta v$ uncertainty in velocity
hPlanck's constant

Worked example

The uncertainty in the position of an electron is

1 \times 10^{-10}\,\text{m}

. What is the minimum uncertainty in its momentum? (

h = 6.63 \times 10^{-34}\,\text{J s}

)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q81Easy

Example 2 · Structure of Atom · Quantum Mechanical Model, de Broglie, Heisenberg and Quantum Numbers

"It is impossible to determine simultaneously the exact position and exact momentum of an electron." This statement is called

It is a fundamental limit, not an instrument error

The uncertainty is built into nature — it is not due to imperfect instruments. Even a perfect measuring device cannot beat the

\dfrac{h}{4\pi}

bound.

Don't confuse it with Pauli or Aufbau

Position-and-momentum together = Heisenberg. Pauli's exclusion principle is about no two electrons having the same four quantum numbers; Aufbau is about the order of filling. The bank swaps these as distractors.

Drill 1 more on heisenberg's uncertainty principle

Concept 3 of 5: The four quantum numbers

Four quantum numbers are the electron's full address: n names the shell (floor), l names the subshell/shape (room type), m_l names the particular orbital (room number) and m_s names the spin (which way it faces). The bank's staple is turning an (n, l) pair into an orbital label — l = 2 always means a d orbital.

Definition

The four quantum numbers and what each fixes:

Principal (n) — the shell / energy level and size; $n = 1, 2, 3, \ldots$ (K, L, M, N).
Azimuthal (l) — the subshell and orbital shape; $l = 0$ to $n-1$ , coded $0 = s, 1 = p, 2 = d, 3 = f$ .
Magnetic (m_l) — the orbital's orientation in space; integers from $-l$ to $+l$ , giving $2l+1$ orbitals per subshell.
Spin (m_s) — the electron's spin direction, $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$ .

To name an orbital, write the value of $n$ then the letter for $l$ : $n=3, l=2 \Rightarrow$ 3d; $n=4, l=3 \Rightarrow$ 4f.

Quantum number	Symbol	What it describes	Allowed values
Principal	n	Shell / main energy level and size of the orbital	1, 2, 3, ... (positive integers)
Azimuthal (subsidiary)	l	Subshell and shape of the orbital (s, p, d, f)	0 to (n-1); coded 0=s, 1=p, 2=d, 3=f l runs only from 0 up to n-1. For n=3, l can be 0, 1 or 2 — never 3.
Magnetic	m_l	Orientation of the orbital in space (which orbital)	-l to +l, i.e. (2l+1) values
Spin	m_s	Direction of the electron's spin	+1/2 or -1/2 only

l fixes the shape (s/p/d/f); the orbital label is n followed by that letter.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q53Easy

Example 3 · Structure of Atom · Quantum Mechanical Model, de Broglie, Heisenberg and Quantum Numbers

Which of the following orbitals is represented by

n = 3

and

l = 2

l ranges from 0 to n-1

The azimuthal quantum number cannot equal or exceed

n

. For

n = 3

the allowed

l

values are 0, 1, 2 only — there is no 3f (that would need

n \geq 4

m_l ranges from -l to +l

The magnetic quantum number runs over the

2l+1

integers from

-l

+l

, including 0. A p subshell (

l = 1

) has

m_l = -1, 0, +1

— three orbitals, not two.

Drill 2 more on the four quantum numbers

Concept 4 of 5: Orbital shapes from l

The azimuthal quantum number l fixes the shape of the orbital: s is a sphere, p is a dumbbell, d is a four-lobed clover leaf. The bank's favourite catch is that four of the five d orbitals look the same but the fifth,

d_{z^2}

, is a different shape — a dumbbell wrapped in a doughnut.

Definition

Shape of the orbital for each value of $l$ :

$l = 0$ (s) — spherical, symmetric about the nucleus; one orbital.
$l = 1$ (p) — dumbbell (two lobes) along an axis; three orbitals $p_x, p_y, p_z$ .
$l = 2$ (d) — mostly double-dumbbell / clover-leaf (four lobes); five orbitals.
$l = 3$ (f) — complex multi-lobed shapes; seven orbitals.

Among the d orbitals, $d_{xy}, d_{yz}, d_{xz}$ and $d_{x^2-y^2}$ are the four-lobed clover leaves, while $d_{z^2}$ is the odd one out — two lobes on the z-axis plus a ring in the xy-plane.

l value	Subshell	Shape	Orbitals in subshell
0	s	Spherical	1
1	p	Dumbbell (two lobes)	3
2	d	Four-lobed clover leaf (except d(z2))	5 d(z2) is the exception: two lobes along z plus a doughnut ring in the xy-plane — a different shape from the other four.
3	f	Complex multi-lobed	7

Number of orbitals in a subshell is 2l+1: s=1, p=3, d=5, f=7.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q70Easy

Example 4 · Structure of Atom · Quantum Mechanical Model, de Broglie, Heisenberg and Quantum Numbers

Which from following d-orbitals has different shape as compared with others?

d(z2) is the shape exception

When asked which d orbital has a different shape, the answer is

d_{z^2}

. The other four (

d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}

) are all four-lobed clover leaves.

Concept 5 of 5: Shell capacity, orbital energy order and nodes

A shell of number n holds

n^2

orbitals and up to

2n^2

electrons — that is the whole answer to 'how many orbitals/electrons in the M shell?'. For energy order the trick is the (n+l) rule: lower

n+l

means lower energy, and a tie is broken by the smaller n.

Definition

Counting and ordering rules:

Orbitals in a shell $= n^2$ ; maximum electrons $= 2n^2$ . (M shell, $n=3$ : 9 orbitals, 18 electrons.)
Electrons in a subshell $= 2(2l+1)$ : s holds 2, p holds 6, d holds 10, f holds 14.
(n+l) rule (Aufbau): the orbital with the lower $n+l$ has lower energy; if two orbitals have the same $n+l$ , the one with the smaller n is lower.
Degeneracy in hydrogen only: for the H atom, energy depends on $n$ alone, so 2s and 2p (same n) are degenerate. In multi-electron atoms they are not.
Nodes: total nodes $= n-1$ ; angular nodes $= l$ ; radial nodes $= n-l-1$ .

Shell capacity, subshell capacity and nodes

\text{orbitals} = n^2,\quad e^- _{\max} = 2n^2,\quad e^-_{\text{subshell}} = 2(2l+1) \qquad \text{nodes} = n-1,\ \text{radial} = n-l-1

nprincipal quantum number (shell)
lazimuthal quantum number (subshell)
n^2number of orbitals in the shell
2n^2maximum electrons in the shell

Worked example

How many orbitals and how many electrons at most can the M shell hold?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q73Easy

Example 5 · Structure of Atom · Quantum Mechanical Model, de Broglie, Heisenberg and Quantum Numbers

Find the number of orbitals and maximum electrons respectively present in M shell?

Break an (n+l) tie with the smaller n

When two orbitals share the same

n+l

(e.g. 3d and 4p both give 5), the one with the smaller n has the lower energy: 3d is below 4p. Only after comparing

n+l

do you look at

n

Degeneracy of 2s and 2p is a hydrogen-only fact

In the hydrogen atom, energy depends only on

n

, so 2s and 2p are degenerate. In any multi-electron atom the (n+l) rule splits them — 2s is below 2p. The bank's degeneracy question is specifically about hydrogen.

Drill 4 more on shell capacity, orbital energy order and nodes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

de Broglie wavelength — wave-particle duality
de Broglie wavelength and momentum
$\lambda = \dfrac{h}{mv} = \dfrac{h}{p} \qquad p = \dfrac{h}{\lambda}$
Heisenberg's uncertainty principle
Heisenberg uncertainty relation
$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} \qquad \Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}$
Shell capacity, orbital energy order and nodes
Shell capacity, subshell capacity and nodes
$\text{orbitals} = n^2,\quad e^- _{\max} = 2n^2,\quad e^-_{\text{subshell}} = 2(2l+1) \qquad \text{nodes} = n-1,\ \text{radial} = n-l-1$

Reference tables (2)

The four quantum numbers4 rows

Quantum number	Symbol	What it describes	Allowed values
Principal	n	Shell / main energy level and size of the orbital	1, 2, 3, ... (positive integers)
Azimuthal (subsidiary)	l	Subshell and shape of the orbital (s, p, d, f)	0 to (n-1); coded 0=s, 1=p, 2=d, 3=f l runs only from 0 up to n-1. For n=3, l can be 0, 1 or 2 — never 3.
Magnetic	m_l	Orientation of the orbital in space (which orbital)	-l to +l, i.e. (2l+1) values
Spin	m_s	Direction of the electron's spin	+1/2 or -1/2 only

l fixes the shape (s/p/d/f); the orbital label is n followed by that letter.

Orbital shapes from l4 rows

l value	Subshell	Shape	Orbitals in subshell
0	s	Spherical	1
1	p	Dumbbell (two lobes)	3
2	d	Four-lobed clover leaf (except d(z2))	5 d(z2) is the exception: two lobes along z plus a doughnut ring in the xy-plane — a different shape from the other four.
3	f	Complex multi-lobed	7

Number of orbitals in a subshell is 2l+1: s=1, p=3, d=5, f=7.

Watch out for (9)

Divide by momentum, not by mass alone
→ de Broglie wavelength — wave-particle duality
Convert Ångström to metres
→ de Broglie wavelength — wave-particle duality
It is a fundamental limit, not an instrument error
→ Heisenberg's uncertainty principle
Don't confuse it with Pauli or Aufbau
→ Heisenberg's uncertainty principle
l ranges from 0 to n-1
→ The four quantum numbers
m_l ranges from -l to +l
→ The four quantum numbers
d(z2) is the shape exception
→ Orbital shapes from l
Break an (n+l) tie with the smaller n
→ Shell capacity, orbital energy order and nodes
Degeneracy of 2s and 2p is a hydrogen-only fact
→ Shell capacity, orbital energy order and nodes

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