a polished Cesium filament is heated to1500K. At low temperatures the electron C

Question

a polished Cesium filament is heated to1500K. At low temperatures the electron Cesium are confined to energies less than or equal to the Fermi, energy E_r = 153w. The elector energy density in Cesium is 10^28 electrons/m^3. when the Cesium is heated to 1500K what is the approximate density of the electrons the surface which have gained enough energy to overcome the Cesium work function of 1.9eV(e g (E - E_c) = 1.9eV) The filament can be modeled by a one dimensional box of length, L with infinity cohential was. The electron energy are quantized with E_n m^2 E_g. If the temperature of the approachers OCT there be a maximum value of n for an energy state that an electron can be occupy, and if so how would you go about determining n_max if you know the linear electron density (i.e 10^16 electrons/m)? If helium is absorbed by the filament the helium can also be modeled by the one dimensional box with energy E(He)_n = n^2E(He)_0s. If the filament is cooled T - 0 what a values with the He atoms occup?

Explanation / Answer

Given data T = 1500 K, Work function of Cs = 1.96 eV , concentration of electrons n = 10^28 / m3

a) Density of electrons emitted normal to the Cs surface per unit area per unit time is calculated from Richrdson- Dushman equation. n(T) = ( 4x pie x m x k^2 / h^3) x T^2 x e^(-work function /kT) = 0.75 x (1500^2) x exp[ (-1.96x 1.6x 10^-19 ) / (1.38 x 10^ -23 x 1500)] = 4.44 x 10 ^ 24 /m^3

b ) when T is tending towards 0 K all the electrons will occupy upto the fermi level. Then the given concentration can be caculated by integrating by taking the limits between o to Ef of Density of states multiplied with Fermi distribution. At T = 0K F(E) =1 Therfore n = Pie /3 x [ 8*m/h^2]^3/2 Ef^3/2 = 1.683 x 10^-22/m3

c) do the same method mass of He to be taken

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