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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 on theme: "Franck Hertz experiment 6B Tam Fei Ying(22) 6B Tam Fei Ying(22) Pang Sze Man(28) Pang Sze Man(28)"— Presentation transcript:

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

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

4 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.

5 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

6 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.

7  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.

8 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.

9 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.

10  A sharp drop in current at V=(6.7-4.9)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.

11 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.

12 THE END


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