NEET > NCERT Exemplar & Revision Notes for NEET > Solved Examples: Electronic Devices

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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-E_{F}/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/[e^{1.45} +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]×10^{7}

= 397.8×10^{3} 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.**

Here, gm = 3 mA/V

= 3×10^{-3} mho

r_{p} = 20000 ohm

µ = r_{p}×g_{m }

= 20000× 3×10^{-3}

= 60

Since, voltage gain = µ / [1+ (r_{p}/R_{L})]

40 = 60 / [1+ (20000/R_{L})]

So, 1+ (20000/R_{L}) = 60/40

= 3/2

Or, 20000/R_{L} = 3/2 – 1

= ½

So, R_{L} = 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 V _{B} = 5V, then what will be the the maximum value of R so that the voltage is above the knee point ?**

Here,V

Substituting 5 V for V

get,

V

5 = 0.7 + [(10

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.

The document Solved Examples: Electronic Devices | NCERT Exemplar & Revision Notes for NEET is a part of the NEET Course NCERT Exemplar & Revision Notes for NEET.

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