Franck Hertz experiment 6B Tam Fei Ying(22) 6B Tam Fei Ying(22) Pang Sze Man(28) Pang Sze Man(28)

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Presentation transcript:

Franck Hertz experiment 6B Tam Fei Ying(22) 6B Tam Fei Ying(22) Pang Sze Man(28) Pang Sze Man(28)

Glass tube contains mercury vapour at low pressure Hot cathode (emitter) C emits electrons by thermionic emission.

Grid G is at a +ve potential V relative to C The plate (anode) P is at a small -ve potential V relative to G.

Thermoelectrons accelerated by +ve grid potential Most of e- pass through the grid Travel towards to plate Micro-ammeter measures the current I The current I as the accelerating voltage V is increased until V=4.9 where there is a sudden drop. I = a gain Another sudden drop at V=9.8V

Electron collide with mercury atoms lose energy of incident electrons 1.In an elastic collision, Total KE conserved. Mercury atom massive >> electron  Carries always negligible KE  Almost no K.E. loss of electron 2.Inelastic collision some KE lost  converted into the energy inside the mercury atom recoil of the mercury atom is negligible  amount of KE lost by e- = gain in E inside the mercury atom.

 Almost all the mercury atoms are in the ground state.  When K.E. max of C < 4.9eV  When an electron hits a mercury atom, there is no way for it to excite the atom.  All collisions are elastic. (the energy of the electron is not lost to the atom).  The electrons go through the grid with the original energy. The energy is enough to overcome the retarding p.d.(Vr).  The electrons are gain energy.

e- collides with mercury atom enough KE  the atom into 1st excited state. After inelastic collision,this amount of E is not enough to overcome the Vr. the current shows a sharp drop. (line a) The p.d. V for every sharp drop marks an allowed value of energy absorption for the atom. The sharp drop at 6.7V corresponds to the transition indicated by line b.

The values of the p.d. for the transitions (I.e. 4.9V and 6.7V) -  excitation potentials of this atom. The corresponding energies (4.9eV and 6.7eV) are called excitation energies.

 A sharp drop in current at V=( )V=1.8V  At this voltage, enough E to raise the atom from 1st excited state  2nd excited state. (line c)  The drop is not observed because at ordinary T extremely few mercury atoms in 1st excited state.

AAfter a mercury atom has been raised to an excited state   back down in a relatively short time. TThe excess energy can be released by emitting EM radiation.

THE END