Q.20 Obtain an expression for the de-Broglie wavelength associated with an electron accelerated from rest through a potential difference of ‘V’ volts.
Best answer
Answer: \(\lambda = \dfrac{h}{\sqrt{2meV}} \approx \dfrac{1.23}{\sqrt{V}}\) nm
An electron (mass \(m\), charge \(e\)) accelerated from rest through a p.d. \(V\) gains kinetic energy
\[\frac{1}{2}mv^2 = eV \Rightarrow v = \sqrt{\frac{2eV}{m}}\]Its momentum is \(p = mv = \sqrt{2meV}\). The de Broglie wavelength is
\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\]Substituting \(h = 6.63\times10^{-34}\) J s, \(m = 9.1\times10^{-31}\) kg and \(e = 1.6\times10^{-19}\) C:
\[\lambda = \frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\,V}} \approx \frac{1.229\times10^{-9}}{\sqrt{V}}\ \text{m} \approx \frac{1.23}{\sqrt{V}}\ \text{nm}\](valid for non-relativistic speeds; with the more precise constants the familiar value is \(1.227/\sqrt{V}\) nm).
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