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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as ${\mathbit{\lambda }}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{226}}{\mathbf{/}}\sqrt{\mathbf{K}}$, in which K is the electron’s kinetic energy in electron-volts.

The required expression for the de Broglie wavelength is, $\frac{1.226×{10}^{-9}{\text{m.eV}}^{\frac{\text{1}}{\text{2}}}}{\sqrt{K}}$ .

See the step by step solution

## Step 1: Write the given data from the question

The classical equation of momentum is, $p=mv$

The kinetic energy is, $K=\frac{1}{2}m{v}^{2}$

## Step 2: Determine the formulas to prove the de Broglie wavelength λ=1.226/K

The expression to calculate the kinetic energy of the electron is given as follows.

$K=\frac{1}{2}m{v}^{2}$ …(i)

Here, m is the mass of the electron and v is the velocity of electron.

The expression to calculate the de Broglie wavelength is given as follows.

$\lambda =\frac{h}{p}$ …(ii)

Here, h is the plank’s constant and $p$ is the momentum.

## Step 3: Prove the Broglie wavelength λ=1.226/K

Calculate the kinetic energy of the electron,

Substitute $\rho /m$ for v into equation (i).

$\begin{array}{rcl}K& =& \frac{1}{2}m{\left(\frac{p}{m}\right)}^{2}\\ K& =& \frac{1}{2}\frac{{p}^{2}}{m}\\ {p}^{2}& =& 2Km\\ p& =& \sqrt{2Km}\end{array}$

Calculate the de Broglie wavelength.

Substitute$\sqrt{2Km}$ for $p$ into equation (ii).

$\lambda =\frac{h}{\sqrt{2Km}}$

Substitute $6.62×{10}^{-34}\text{Js}$ for h and $9.11×{10}^{-31}\text{kg}$for m into above equation.

Hence the required expression for the de Broglie wavelength is, .