What is the main aim of school laboratory experiment using the gas discharge tube?
What is being discharged? The metal at the cathode?
Why should there be low pressure inside the tube and high voltage supplied across the electrodes? Is high voltage used for thermionic emission?
When we are applying the equation 1/2 m v square = e V, where m is the mass of electron, e is charege of electron and V is voltage supplied, at which position does the velocity v of electron is it refering to? is it the initial velocity of emitted electrons at the cathode? is there any initial kinetic energy at eh cathode?
When the gas in the tube begins to act as a conductor the current which flows carries electric charge between the cathode and anode, "discharging" the tube. The power supply replaces the charge so the discharge does not reduce the voltage completely.
At sufficiently low pressures the number of gas molecules sharing the voltage drop from one end of the tube to the other is reduced to the point that the applied voltage can ionize the gas, causing it to become a conductor. There is no thermionic emission because the cathode is not heated.
The expression eV is the potential energy of an electron where V is the voltage drop through which an electron has fallen. An electron near the cathode has experienced about zero drop in voltage. As it responds to the attraction of the anode and moves away from the cathode, the change in potential, V, at its location increases until near the anode it has fallen through the entire potential across the tube. This decrease in potential energy is matched by an increase of kinetic energy such that 1/2mv^2=eV. The velocity v is the instantaneous velocity at any time and the potential V is the voltage change between the electron's zero velocity position and its current position. The initial energy at the cathode is essentially all potential and not kinetic. At the anode the situation is reversed with the energy essentially all kinetic and no potential.