12.Determination of Fermi energy

AIM: To determine the Fermi energy of the copper.
APPARATUS: DC Regulated power supply, Milliammeter, Voltmeter, Copper wire and Screw gauge.
PRINCIPLE: The energy of the highest occupied level by an electron at absolute zero temperature is called the Fermi energy. Fermi energy is given by,
EF = ½ mvF2
Where ‘m’ is the mass of the electron and ‘vF’ the Fermi velocity.
But vF = lF/t where lF is mean free path and t the relaxation time.
Since conductivity s = 1/r = L/RA = ne2t/m,
t = mL/ne2RA = mL/ne2Rpr2 [A=pr2 ]
\vF = ne2Rpr2 lF/ mL
Now EF = ½ mvF2 = ½ m (ne2Rpr2 lF/ mL)2
For Copper, n = 8.5 x 1028 /m3
lF = 53 x 10-9 m
e = 1.6 x 10-19 C
m = 9.1 x 10-31 Kg
\ EF = 7.16 x 10-2 x (r4/L2) x R2
But R = slope of the Voltage-Current graph.
\ EF = 7.16 x 10-2 x (r4/L2) x [Slope]2

PROCEDURE: The circuit arrangement is made as shown in figure. The voltage is varied gradually and the corresponding current values are noted down. A graph can be plotted with the voltage along Y-axis and the current along the X-axis. The slope of the graph gives the resistance of the given copper material. The radius of the copper wire is measured using a screw gauge and length by using a meter scale.

OBSERVATIONS:
Radius of the wire (r) =
Length of the wire (L) =
Voltage (V) Current (I)

CALCULATIONS:
Slope of the graph = AB/BC =
EF = 7.16 x 10-2 x (r4/L2) x [Slope]2 =
RESULT: Fermi energy of the given material =