Describe the Frank-Hertz experiment. Discuss its result and outline limitations.

This answer is restricted. Please login to view the answer of this question.

Login Now

Franck-Hertz Experiment was used to describe the quantum nature of atoms clearly. The existence of discrete energy levels in atoms was demonstrated directly by James Franck and Crustav Hertz in 1914.

A scheme of the experimental set up shown in figure (1)

Franck Hertz Experiment | CSIT Guide

Figure 1: Franck-Hertz Experiment.

The tube contains three electrodes: a filament (F), a plate (P), and a grid (G). A grid is a charged screen that can attract or repel electrons. But due to the open space, the majority of the electron passes through it. A variable accelerating potential Vo is applied between the filament and the grid and a small retarding voltage Vr(≈1V) is applied between the plate and the grid. The electrons will reach the grid with a kinetic energy Ek = eVo. After reaching the grid majority,  these electrons will go through the open space and contribute to the plate current I.

If Vr>Vo, the electrons will be turned back before they can reach the plate and they will not contribute to the current measured by Ammeter.

But even if Vr<Vo, the electron will not able to reach the plate if they lose enough kinetic energy through collision with atoms in the tube as they travel between the filament and the grid.

current vs voltage | CSIT Guide

Figure 2: Dependence of plate current with Vo

Consider the case of beam of slow electrons traveling through the mercury vapor at low pressure. If the electron suffers no energy loss due to elastic collision with the atoms of the gas. They will reach the collector plate, increasing the accelerating potential, collector current increases. As ‘V’ reaches the value of critical potential 4.9V, an electron acquires 4.9V of energy on reaching ‘G’. The electron losses all its energy in an in-elastic collision with the mercury atom, so the electrons unable to reach plate ‘p’ and current decreases. On increasing the accelerating potential beyond 4.9V again the electrons reach to the plate at V=9.8V, the current again dip to a second minimum. This can be explained if an electron of 9.8V energy suffers to consecutive in-elastic collision with the different mercury atom before reaching the plate. Such an electron excited both the mercury atoms to their first excited state losing 4.9eV energy in each collision. This explains the second minimum.

Each time there is an in-elastic collision, the mercury atoms will be excited and return to the ground state by emission of photons. There should be a spectral line whose frequency is given by hf = 4.9eV or λ = 2530 Α°. Such a wave length is found in the spectrum of Hg.

An energy diagram for Hg from spectral data in shown in figure(3)

User Loaded Image | CSIT Guide


The energy difference ΔE between the first excited state and ground state is

ΔE = 10.4eV -5.5eV = 4.9 eV.

The energy difference ΔE between the second excited state and ground state is ΔE = 10.4eV – 3.7eV = 6.7 eV. But we may not observe a dip when Vo=6.7V, This is because the number of electron that are able to avoid the 4.9eV collision and gather 6.7eV is a small fraction of the total number of electron. Which reduces the magnitude of 6.7eV dip and makes it more difficult to observe. This can be observed if the sensitivity of the experiment is increased.

Thus, this experiment shows that the energy lost by the electron in its inelastic collision with the mercury atom reappears as a quantum of energy of wavelength hc/E. This experiment shows in a very convincing way the existence of discrete energy levels in the mercury atom. While it isn’t evident in the original measurements of the figure. this series of dips in current at approximately 4.9 volt increments continues to potentials of at least 70 volts.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

  Loading . . .