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Solved Examples on Solid and Electronic Device

Question 1:-

(a) What is the probability that a quantum state whose energy is 0.10eV above the Fermi energy will be occupied? Assume a sample temperature of 800 K. 

(b) What is the probability of occupancy for a state that is 0.10 eV below the Fermi energy?

Solution:-

(a) We can find P(E) from equation P(E) = 1/[e(E-Ef)/kT+1]. But let us first calculate the (dimensionless) exponent in that equation:

E-EF/kT = (0.10 eV)/(8.62×10-5 eV/K) (800 K)

             = 1.45

Inserting this exponent into equation P(E) = 1/[e(E-Ef)/kT+1], we get,

P(E) = 1/[e1.45 +1]

       = 0.19 or 19%

Therefore, the  probability that a quantum state whose energy is 0.10eV above the fermi energy occupied will be19%.  

(b) The exponent in equation P(E) = 1/[e(E-Ef)/kT+1] has the same absolute value as in (a) but is now negative. Thus from this equation,

P(E) = 1/e-1.45+1

       = 0.81 or 81%

From the above observation we conclude that, the probability of occupancy for a state that is 0.10 eV below the Fermi energy would be 81%. For states below the Fermi energy, we are often more interested in the probability that the state is not occupied. This just 1- P(E), or 19%. Note that it is the same as the probability of occupancy in (a).

 

Question 2:-

Calculate the frequency of radio-waves radiated out by an oscillating circuit consisting of a capacitor of capacity 0.02 µF and inductance 8µH.

Solution:-

Here, C = 0.02 µF 

           = 2×10-8 F

         L = 8 µH

           = 8×10-6 H

We know that, frequency, f = 1/[2π√LC]

 

Substitute the value of L nad C in the equation  f = 1/[2π√LC], we get,

 f = 1/[2π√LC]

  = 1/[2π√(2×10-8 F) (8×10-6 H)]

 = [1/2π×4]×107

= 397.8×103 Hz

= 397.8 kilo Hz

From the above observation we conclude that, the frequency of radio-waves radiated out by an oscillating circuit consisting of a capacitor of capacity 0.02 µF and inductance 8µH would be 397.8 kilo Hz. 

 

Question 3:-

A triode has a mutual conductance of 3 mAV-1 and anode resistance of 20000 ohm. Find the load resistance which must be introduced in the circut to obtain a voltage gain of 40.

Solution:- 

Here, gm = 3 mA/V 

               = 3×10-3 mho

         rp = 20000 ohm

        µ = rp×gm 

          = 20000× 3×10-3

             = 60

Since, voltage gain = µ / [1+ (rp/RL)]

                          40 = 60 / [1+ (20000/RL)] 

So, 1+ (20000/RL) = 60/40

                            = 3/2

Or, 20000/R= 3/2 – 1

         ...

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