Question

An electron of mass m and charge e is accelerated from rest through a potential difference V. Prove that, after acceleration, the de Broglie wavelength associated with the electron is proportional to \(\dfrac{1}{\sqrt{V}}\).

Hint:

Equate the kinetic energy to eV, find the momentum \(p = \sqrt{2meV}\), then put it in \(\lambda = h/p\).

Answer 1

Step 1: Understanding the Concept:
A moving particle has a wavelength given by de Broglie relation \(\lambda = h/p\), where \(p\) is its momentum and \(h\) is the Planck constant.
We must find \(p\) of the electron after it falls through a potential difference \(V\).

Step 2: Find the speed from energy conservation:
The electron starts from rest. The work done by the field is \(eV\), and it appears as kinetic energy.
\[ \frac{1}{2} m v^2 = eV \quad \Rightarrow \quad v = \sqrt{\frac{2eV}{m}} \]

Step 3: Find the momentum:
\[ p = m v = m \sqrt{\frac{2eV}{m}} = \sqrt{2 m e V} \]

Step 4: Write the wavelength:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}} = \frac{h}{\sqrt{2 m e}} \cdot \frac{1}{\sqrt{V}} \]
Here \(h\), \(m\) and \(e\) are constants, so the first factor is a constant.

Final Answer:
The de Broglie wavelength of the accelerated electron is \(\lambda = h / \sqrt{2meV}\), so it is proportional to \(1/\sqrt{V}\). Hence proved. \[ \boxed{\lambda \propto \frac{1}{\sqrt{V}}} \]