# Davisson and Germer Experiment | Construction, Working

## Davisson and Germer Experiment

The wave nature of electrons was first experimentally verified independently by C. J. Davisson and L. H. Germer in 1927 and by G. P. Thomson in 1928 while observing diffraction effects with beams of electrons scattered by crystals. The experimental arrangement is schematically shown in figure.

### Construction

It has an electron gun made up of a tungsten filament F, heated by a low voltage battery and the filament is coated with barium oxide. Emitted electrons from filament are accelerated to a desired velocity by applying required potential/voltage from a high-voltage power supply. C is a hollow metallic cylinder with a hole along the axis and is kept at negative potential to get a convergent beam of electrons emitted from filament. It acts as a cathode. A is a cylinder with fine hole along its axis acting as an anode.

The cathode and anode form an electron gun by which a fine beam of electrons can be obtained of different velocities by applying different accelerating potentials. N is a nickel crystal cut along cubical diagonal, D is an electron detector which can be rotated on a circular scale and is connected to a sensitive galvanometer which records the current.

### Working

From electron gun a fine beam of accelerated electrons is made to fall normally on the surface of nickel crystal. The atoms of the crystal scatter the incident electrons in different directions. The detector detects the intensity of the electron beam scattered in particular direction by rotating the electron detector on circular scale at different positions.

According to de Broglie hypothesis, the wavelength of the wave associated with electron is given by

\lambda=\frac{h}{mv} .....(1)

Suppose an electron is accelerated by a potential difference V and the charge of an electron is e, then the kinetic energy acquired by the electron will be eV. If the mass of the electron is m and its velocity is V, then the kinetic energy of electron,

\frac{1}{2}mv^2=eV

v=\sqrt{\frac{2eV}{m}}

\lambda=\frac{h}{m\sqrt{\frac{2eV}{m}}}

\lambda=\frac{h}{\sqrt{2meV}} .....(2)

Putting the value of constant e = 1.6x10-19 C, m = 9.1x10-31 kg, h = 6.626x10-34 Js in the above eq. (2), we get

\lambda=\frac{6.626\times 10^{-34}}{sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}}}

\lambda=\frac{12.27}{sqrt{V}}.Ã…