de Broglie Wavelength

PHYMatter (de Broglie) Waves

The de Broglie hypothesis assigns a wavelength $\lambda=h/p$ to every moving particle, unifying the wave and particle pictures of matter. For charged particles accelerated through a potential difference, this gives compact working formulas that ISC examines through repeated numericals on electrons, protons and alpha particles.

The headline insight — that macroscopic objects have unmeasurably tiny wavelengths while electrons have atom-scale wavelengths — is itself a frequent comment-on-your-answer question.

de Broglie wavelength

\lambda = \dfrac{h}{p} = \dfrac{h}{m v}

\lambda

= wavelength (m),

h=6.63\times10^{-34}

J s,

p=mv

= momentum,

m

= mass (kg),

v

= speed (m/s).

In terms of kinetic energy

\lambda = \dfrac{h}{\sqrt{2 m K}}

K

= kinetic energy (J). Convert

K

from eV using

1\,\text{eV}=1.6\times10^{-19}

J before substituting.

Accelerated charged particle

\lambda = \dfrac{h}{\sqrt{2 m q V}}

q

= charge of particle,

V

= accelerating potential difference (V). Here

K=qV

. For an alpha particle

q=2e

Electron shortcut

\lambda = \dfrac{12.27}{\sqrt{V}}\ \text{\AA}

V

= accelerating voltage in volts; valid for non-relativistic electrons. Gives

\lambda

directly in \aa ngstr\"om.

Equal kinetic energy ratio

\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2}{m_1}}

For two particles of the same kinetic energy;

m_1,m_2

their masses. Lighter particle has the longer wavelength.

The relation $\lambda=h/p$ applies to all matter — electrons, protons, atoms and even a cricket ball — but the wavelength is detectable only when it is comparable to the size of the apertures or atomic spacings available.
Choose the formula by what is given: speed $\to h/mv$ ; kinetic energy $\to h/\sqrt{2mK}$ ; accelerating voltage $\to h/\sqrt{2mqV}$ .
For an alpha particle use mass $\approx 6.64\times10^{-27}$ kg and charge $q=2e=3.2\times10^{-19}$ C; the factor of 2 in the charge is easy to drop.
At equal kinetic energy, $\lambda \propto 1/\sqrt{m}$ , so an electron has a far longer de Broglie wavelength than a proton of the same energy.
A 150 g cricket ball at 40 m/s has $\lambda \sim 10^{-34}$ m — utterly negligible, which is why wave behaviour is never observed for everyday objects; always add this comment when asked.
These formulas are non-relativistic; they are accurate as long as the particle speed is well below $c$ , which holds for ISC-level voltages.
Keep all SI units consistent (mass in kg, energy in J) and your answer comes out in metres; convert to \aa ngstr\"om ( $1\,\text{\AA}=10^{-10}$ m) or nm only at the end.

Where the marks go

Forgetting to convert kinetic energy from eV to joule before using $\lambda=h/\sqrt{2mK}$ — a factor-of- $1.6\times10^{-19}$ error.
Using charge $e$ instead of $2e$ for an alpha particle in $\lambda=h/\sqrt{2mqV}$ .
Applying the shortcut $\lambda=12.27/\sqrt{V}\,\text{\AA}$ to protons or alpha particles — it is derived specifically for electrons.
Inverting the mass ratio: at equal $K$ the lighter particle has the larger $\lambda$ , so the electron-to-proton wavelength ratio is $\sqrt{m_p/m_e}>1$ .

How the board asks it

Numericalaccelerated charged particle, $\lambda=h/\sqrt{2meV}$
An electron is accelerated from rest through a potential difference of $100\,\text{V}$ . Calculate the de Broglie wavelength associated with it. (Given $m_e=9.1\times10^{-31}\,\text{kg}$ , $e=1.6\times10^{-19}\,\text{C}$ , $h=6.63\times10^{-34}\,\text{J s}$ .)
Define / statede Broglie hypothesis $\lambda=h/p$
State de Broglie's hypothesis and write the expression for the de Broglie wavelength of a particle of mass $m$ moving with velocity $v$ .
Derive / prove $\lambda=h/\sqrt{2meV}$ from $\lambda=h/p$
Derive an expression for the de Broglie wavelength of an electron accelerated from rest through a potential difference $V$ , and hence show that $\lambda=\dfrac{12.27}{\sqrt{V}}\,\text{\AA}$ .
Numericalequal kinetic energy, $\lambda\propto 1/\sqrt{m}$
An electron and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths $\lambda_e/\lambda_p$ . (Given $m_p/m_e=1836$ .)
Give reasonswave detectability when $\lambda$ is comparable to aperture size
A cricket ball and an electron move with the same speed, yet wave behaviour is observed only for the electron. Give reasons.
Applicationalpha-particle mass and charge $q=2e$
An alpha particle and a proton are accelerated from rest through the same potential difference. Compare their de Broglie wavelengths, and justify your answer.

Practise this topic

PreviousPhoton Energy, Momentum & Number NextMultiple Choice & Assertion-Reason

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.