JEE Mains Chemistry · Structure of Atom

de Broglie Waves and the Uncertainty Principle

A moving particle has a wavelength h/mv, a Bohr orbit holds a whole number of these waves, and the position and momentum of a particle can never both be known exactly.

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Why this matters

Fifteen PYQs, eleven of them numerical-answer — the highest share in the chapter. Six find a de Broglie wavelength from mass and speed, kinetic energy or an accelerating voltage; four fit whole waves round a Bohr orbit; five apply Δx·Δp ≥ h/4π. Three ideas cover the page.

Concept 1 of 3: de Broglie wavelength

Every moving particle behaves as a wave of wavelength

h/p

. Only momentum matters: two particles with equal momentum have equal wavelength, whatever their masses. For a heavy object the wavelength is far too small to notice. For an electron it is about the size of an atom.

Definition

$\lambda=\dfrac{h}{mv}=\dfrac{h}{p}$ .
From kinetic energy: $p=\sqrt{2mK}$ , so $\lambda=\dfrac{h}{\sqrt{2mK}}$ .
A charge $q$ accelerated through $V$ : $K=qV$ , so $\lambda=\dfrac{h}{\sqrt{2mqV}}$ .
$\lambda$ against $p$ is a rectangular hyperbola; $\lambda$ against $1/p$ is a straight line through the origin.
Cathode rays are electrons. They travel from cathode to anode, and their properties do not depend on the electrode material or the gas.

de Broglie relation

\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}

Worked example

An electron is accelerated from rest through 150 V. Find its de Broglie wavelength. (

m=9.11\times10^{-31}\ \mathrm{kg}

e=1.602\times10^{-19}\ \mathrm{C}

h=6.626\times10^{-34}\ \mathrm{J\,s}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q53Moderate

Example 1 · Structure of Atom · de Broglie Waves and the Uncertainty Principle

The wavelength of an electron and a neutron will become equal when the velocity of the electron is

x

times the velocity of neutron. The value of

x

is. (Nearest Integer) (Mass of electron is

9.1 \times10^{- 31}\text{ }kg

and mass of neutron is

1.6 \times10^{- 27}\text{ }kg

)

Equal wavelength means equal momentum

Equal

\lambda

gives

m_1v_1=m_2v_2

, not equal speeds. The lighter particle must move faster by the mass ratio.

Work in kg and J

Masses in grams or amu, and energies in eV, must be converted first.

1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}

1\ \mathrm{amu}=1.66\times10^{-27}\ \mathrm{kg}

Drill 5 more on de broglie wavelength

Concept 2 of 3: de Broglie waves in a Bohr orbit

Bohr's rule

mvr=nh/2\pi

says the same thing as: exactly

n

electron waves fit round the

n

-th orbit. So the wavelength is the circumference divided by

n

. Since the radius goes as

n^2

, the wavelength goes as

n

Definition

$2\pi r_n=n\lambda$ .
With $r_n=\dfrac{n^2a_0}{Z}$ : $\lambda_n=\dfrac{2\pi na_0}{Z}$ .
Hydrogen: $\lambda_1=2\pi a_0$ , $\lambda_2=4\pi a_0$ , $\lambda_3=6\pi a_0$ .
Frequency of the electron wave: $\nu=\dfrac{v}{\lambda}=\dfrac{mv^2}{h}=\dfrac{2KE}{h}$ .

Waves in the n-th orbit

\lambda_n=\frac{2\pi r_n}{n}=\frac{2\pi n a_0}{Z}

Worked example

Find the de Broglie wavelength of the electron in the second orbit of

\mathrm{He^+}

, in terms of

a_0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q140Moderate

Example 2 · Structure of Atom · de Broglie Waves and the Uncertainty Principle

If the radius of the first orbit of hydrogen atom

a_{0}

, then de Broglie's wavelength of electron in

3^{\text{rd~}}

orbit is

The wavelength grows as n, not n²

The radius goes as

n^2

, but

n

waves share the circumference. So

\lambda\propto n

: the fourth orbit of H has

\lambda=8\pi a_0

, not

32\pi a_0

Drill 3 more on de broglie waves in a bohr orbit

Concept 3 of 3: Heisenberg's uncertainty principle

The more tightly you pin down where a particle is, the less you know about its momentum. The product of the two uncertainties has a floor of

h/4\pi

. For an electron in a tiny space, the uncertainty in speed becomes huge. For a ball, it is far too small to matter.

Definition

$\Delta x\cdot\Delta p\ge\dfrac{h}{4\pi}$ , and $\Delta p=m\Delta v$ .
Minimum uncertainty in speed: $\Delta v=\dfrac{h}{4\pi m\Delta x}$ .
If $\Delta x=\Delta p$ : $\Delta p=\sqrt{\dfrac{h}{4\pi}}$ , so $\Delta v=\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}$ .
A particle confined to a region takes that region's size as $\Delta x$ .
It rules out a fixed electron path, so it rules out Bohr's orbits.

Uncertainty principle

\Delta x\cdot m\Delta v\ge\frac{h}{4\pi}

Worked example

An electron is confined to a region

1.0\times10^{-10}\ \mathrm{m}

wide, about the size of an atom. Find the minimum uncertainty in its speed. (

m=9.11\times10^{-31}\ \mathrm{kg}

h=6.63\times10^{-34}\ \mathrm{J\,s}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q147Moderate

Example 3 · Structure of Atom · de Broglie Waves and the Uncertainty Principle

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter

10^{- 15}\text{ }m

is _____

\times10^{9}{\text{ }ms}^{- 1}

(nearest integer) [Given: mass of electron

= 9.1 \times10^{- 31}\text{ }kg

, Plank's constant

(h) = 6.626 \times10^{- 34}Js

] (Value of

\pi = 3.14

)

Mass in kg

h

is in J s, so the mass must be in kg. If a question asks for the mass in grams, find it in kg first, then multiply by 1000.

Δx = Δp is not Δx = Δv

Equal uncertainties in position and momentum give

\Delta v=\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}

. Setting

\Delta x=\Delta v

instead gives a different, wrong answer.

Drill 4 more on heisenberg's uncertainty principle

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
de Broglie relation
$\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$
de Broglie waves in a Bohr orbit
Waves in the n-th orbit
$\lambda_n=\frac{2\pi r_n}{n}=\frac{2\pi n a_0}{Z}$
Heisenberg's uncertainty principle
Uncertainty principle
$\Delta x\cdot m\Delta v\ge\frac{h}{4\pi}$

Watch out for (5)

Equal wavelength means equal momentum
→ de Broglie wavelength
Work in kg and J
→ de Broglie wavelength
The wavelength grows as n, not n²
→ de Broglie waves in a Bohr orbit
Mass in kg
→ Heisenberg's uncertainty principle
Δx = Δp is not Δx = Δv
→ Heisenberg's uncertainty principle

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