Question

In electron beam lithography patterns are exposed with electrons. The small wavelength of electrons helps to achieve small feature sizes (in order of nm). Assume an electron gun of energy 100 keV. What is the wavelength of these electrons?
Assume the mass of electron =9.109×10−31 kg and Planck's constant, h=6.626×10−34 J.s. Neglect the relativistic effect.

Answer 1

We are given an interesting practical application of Wave Nature of Electron.

We have the energy of the electron, and we have to calculate the corresponding wavelength.

Wavelength of Matter Particles is given by the De-Broglie Formula. That is:

$\boxed{\lambda=\frac{h}{p}}$

Where p is the Momentum.

But here we don't have momentum. We only have the Kinetic Energy of the Electron. So we need a relation between Kinetic Energy and Momentum.

We have:

$\displaystyle K=\frac{1}{2}mv^2 \\ \\ \\ \text{Multiplying and dividing by m} \\ \\ \\ \implies K=\frac{1}{2}mv^2 \times \frac{m}{m} \\ \\ \\ \implies K = \frac{1}{2m} \times (mv)^2 \\ \\ \\ \implies K=\frac{1}{2m} \times p^2 \\ \\ \\ \implies p^2=2mK \\ \\ \\ \implies \boxed{p=\sqrt{2mK}}$

We can put this in the De-Broglie Equation, so we have:

$\displaystyle \lambda = \frac{h}{p} \\ \\ \\ \implies \boxed{\lambda=\frac{h}{\sqrt{2mK}}}$

Our data is:

$h=6.626\times 10^{-34} \, \, Js \\ \\ m=9.109\times 10^{-31} \, \, kg \\ \\ K=100 \, \, keV=10^5 \, \, eV$

We also know that:

$1 \, \, eV = 1.602 \times 10^{-19} \, \, J$

So,

$K=10^5 \, \, eV = 10^5 \times 1.602 \times 10^{-19} \, \, J =1.602 \times 10^{-14} \, \, J$

We can finally find wavelength:

$\displaystyle \lambda=\frac{h}{\sqrt{2mK}}\\\\\\\implies\lambda=\frac{6.626\times 10^{-34}}{\sqrt{2\times 9.109\times 10^{-31}\times 1.602\times 10^{-14}}}\\\\\\\\\implies \huge \boxed{\bold{\lambda \approx 3.878 \times 10^{-12} \, \, m}}$