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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 be 19%.  
(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×g
= 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/RL = 3/2 – 1
= ½
So, RL = 40000 ohm
Thus, from the above observation we conclude that,  the load resistance which introduced in the circut to obtain a voltage gain of 40 would be 40000 ohm.

Question 4: The junction diode in the following circuit requires a minimum current of 1mA to be above the knee point (0.7V) of its I-V characteristic curve. The voltage across the diode is independent of current above the knee point, If VB = 5V, then what will be the the maximum value of R so that the voltage is above the knee point ?
Solution:
Here,VB = Vknee + IR
Substituting 5 V for VB, 0.7 V for Vknee and 10-3 A for I in the equation VB = Vknee + IR, we
get,
VB = Vknee + IR
5 = 0.7 + [(10-3) (R)]
Or, R = 4.3 KΩ
From the above observation we conclude that, the the maximum value of R so that the voltage is above the knee point would be 4.3 KΩ.

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