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QUESTION: 1

Chromium has BCC structure. Its atomic radius is 0.1249 nm. The free volume/ unit cell is

Solution:

Given data are

Atomic radius of chromium, r = 0.1249 nm.

Free volume/unit cell = ?

If 'a' is the BCC unit cell edge length, then the relation between a and 'r' is

Volume of unit cell. V = a^{3} = (0.28845)^{3} nm^{3}

= 0.024 nm^{3 }

Number of atoms in BCC unit cell = 2

Hence volum e o f a tom s in unit cell,

Free volume/unit cell = V - v = 0.00767 nm^{3}

QUESTION: 2

The separation of a {1 2 3} plane of an orthorhombic unit cell with a = 0 .8 2 nm, b = 0.94 and c = .75nm. is

Solution:

The separation of hkl plane is given as for an orthorhombic lattice s

QUESTION: 3

Consider a BCC crystal with lattice constant a’. Determine the no. of atoms per unit area in the ( 1 .1 . 1) plane if a = 1A^{0}

Solution:

No. of atoms at vertex in (1. 1. 1) plane =

No. of atoms at centre = 1

Total no. of atoms

Area of plane

No. of atoms per unit area in the 1 .1 . 1 plane =

QUESTION: 4

What is the ratio of the nearest neighbour distance to the next nearest neighbour distance in a simple cubic crystal?

Solution:

We know, in a simple cubic lattice all atoms are at eight corners of cubic lattice of length a.

So, In a simple cubic lattice, the distance of nearest neighbour d_{1} = a and the distance of next nearest neighbour d_{1} = a√2

QUESTION: 5

What is the maximum radius of the sphere that can just fit into the void at the body centre of the fcc structure coordinated by the facial atom. Given r is the radius of atom

Solution:

The situation is shown in Fig. Let R be the radius of the sphere that can just fit into the void.

From figure,

4 + R = a/2

∴ R = (a/2) - r ..(1)

We know that for fee structure

a = 4 r / √2 ...(2)

Substituting the value of a from eq.(2) in eq. (1), we get

QUESTION: 6

Xenon crystallizes in fee lattice and the edge of the unit cell is 620 pm. then the radius of xenon atom is

Solution:

For fee lattice 4r = √2 * a. where a = 620 pm

or r = 1/2 √2 a

1/2 √2 x 620 pm = 219.20 pm

QUESTION: 7

Determine the M. I. of a plane that makes intercepts of on the co-ordinate axes of an orthorhombic crystal with a:b:c = 4:3:2

Solution:

Here the unittranslations are a = 4, b = 3 andc = 2 following the same procedure

i) Intercepts 2 3 4

ii) Division by unit translation

iii) R eciprocals 2 1 1/2

iv) After clearing fraction 4 2 1

Therefore the Miller indices of the plan is (421)

QUESTION: 8

The rock salt (NaCI) has fcc structure and contains 4 molecules per unit cell. Calculate the lattice constant for the crystal.

Molecular weight of NaCI = 58.45 kg/kmol

Density = 2180 kg/m^{3}

Avogadro Number N_{A} = 6.02 * 10^{26} kmol^{-1}

Solution:

QUESTION: 9

Which of the following output curve is correct for the given circuit.

Solution:

diode D_{1 }is always reverse biased

Then V_{0 }-5V (for the complete cycle)

QUESTION: 10

The current following through (R_{L} = 5kΩ). if the zener diode used in the circuit has rating of 10 V:-

Solution:

Voltage across R_{L} = 10V.

R_{L}=5kΩ

QUESTION: 11

The concentration of holes in Silicon semiconductor incorrectly the represents the variation with the following parameters as :

Solution:

Then we can clearly see that “E_{1}" affects the concentration of holes in a Semiconductor.

QUESTION: 12

Find the ratio of intercepts on the crystal axes by plane (231) in a simple cubic lattice.

Solution:

Let x_{1}. x_{2}, x_{3} be the intercepts by the plane on the axes. In terms of axial units the intercepts are

where m, n. p are numbers.

Now

m : n : p

= 3 : 2 : 6

X_{1 }: X_{2 : }X_{3 }= 3 : 2 : 6

QUESTION: 13

Find the output voltage in gives fig.

Solution:

Determine the output voltage in Fig.

= - (0.9) 10 = - 9V

QUESTION: 14

A zener diode in the circuit shown in the figure below, has a knee current of 5 mA. and a maximum allowed power dissipation of 330 mW. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage V_{0} at 6V?

Solution:

Maximum load current = (60 - 5) m

A = 55mA

Now,

l_{L} min = 60 mA - 50 mA = 10mA

QUESTION: 15

Silicon diode is less suited for low voltage rectifier operation, because

Solution:

Silicon diode is less suited for low voltage rectifier operation because its breakdown voltage is high.

*Multiple options can be correct

QUESTION: 16

In a single state C.E. amplifier using battery supply. V_{cc} = -20 V and the transistor has a minimum β value of 20 and I_{co} = 10 μA. Quiescent point at V_{ce} = 6V and I_{c}= 2mA if R_{c} = 4 kΩ. Choose the correct statement

Solution:

Here

Now

I_{C} = I_{B} + I_{C}

= 0.1mA + 2mA = 2.1 mA

If V_{e} be the voltage across R_{e}. then

Now

Let I_{1}, and l_{2}, be the currents in resistors R_{1}, and R_{2} and l_{2} = 10 I\b

Now

and

Hence

*Multiple options can be correct

QUESTION: 17

For this circuit V_{z} = 10 V, R = 1 kΩ and l_{Zmax} = 2.5 mA. If V_{L} is maintained at 10 V. choose the correct statement :-

Solution:

*Multiple options can be correct

QUESTION: 18

Figure show that a silicon transistor with β = 100 is biased by the resistor method. Which of the following statements are correct?

Solution:

This locates the first point.

When V_{ce} = 0,

This located the second point.

A line joining the above two points is known as d.c. load line as shown in fig. (b)

(ii) Operating point Q.

We know that in a silicon transistor V_{be} = 0.7 volt

∴

∴ Collector current

Now

So operating point is 6 volt 1 mA.

*Multiple options can be correct

QUESTION: 19

An NPN transistor circuit shown in the following figure has

∝ = 0.985

Which of the following statements are correct?

Solution:

Here α = 0.985

We know

∴ β = 66

Base current.

Voltage across R_{2} is given by

Voltage across R_{1} is given by

Current flowing through R_{1} and R_{2}

∴ Resistance, R_{1} =

Voltage across R_{c} = V_{cc} - V_{ce} - Ve

= 20 - 5 - 2 x 2 = 11 volt

Collector resistance,

*Multiple options can be correct

QUESTION: 20

Choose the correct statement for the given output-amp circuit.

Solution:

Voltage at inverting terminal V_{x} = 1 volt (a s non inverting terminal is al 1 V )

V_{p} = 2 volt

At node X :

or V_{Y} = 0

At node P:

or V_{0} = 4 volt.

*Answer can only contain numeric values

QUESTION: 21

Electrons are accelerated to 344 volts and are reflected a crystal. The first reflection maximum occurs when glancing angle is 60^{0}. The spacing of the crystal is_____

Planck's constant, h = 6.62 * 10^{-34} joule-sec.

Charge on election, e = 1.6 * 10^{-19} coul.

mass of the election , m = 9 * 10^{-31} kg

Solution:

The de-Broglie wavelength associated with electrons is given by

(∴ E = e V joule)

∴

According to Bragg’s law.

Substituting the values, we get

= 0.38 x 10^{-10} m = 0.38

*Answer can only contain numeric values

QUESTION: 22

At very low temperature the specific heat of rock salt varies with temperature according to Debye's T^{3} law

θ_{D} for rock salt is 231 K ._____ J heat is required to raise the temperature of 2K mol of rock salt from 10 to 50 K.

Solution:

272000

Heat required to raise the temperature of 2 K mol. of salt through temperature of dT

dQ = mC dT (∴ m = 2 K mol)

Hence total heat required to raise the temperature from 10K to 50K is

substituting the values we have

*Answer can only contain numeric values

QUESTION: 23

Mobilities of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m^{2}V^{-1} S^{-1} and 0.17m^{2}V^{-1}S^{-1} respectively. If the conductivity of the specimen is 2.12 Ω^{-1} m^{-1}, then the forbidden energy gap is _____ eV.

Solution:

0.720

Conductivity of an intrinsic semiconductor is given by

or

= 2.5 x 10^{19} / m^{3}

But

for C = 4.8 x 10^{21}

and k_{B}T = 1.38 x 10^{-23} x 300J =

We have

or

and

E_{g} = 2 x 13.8 x 0.025875 = 0.720 eV.

*Answer can only contain numeric values

QUESTION: 24

A semiconductor has an electron concentration of 0.45 x 10^{12} m^{-3} and a hole concentration of 5.0 x 10^{20} m^{3}. Its conductivity is _________ Sm^{-1}. Given electron mobility = 0.135 m^{2 }V^{-1} s^{-1}; hole mobility = 0.048 m^{2} V^{-1} s^{-1}.

Solution:

The conductivity of a semiconductor is the sum of the conductivities due to electrons and holes and is given by

As per given data n_{e} is negligible as compared to n_{h} so that we can write

where S (seamen) stands for Ω^{-1}.

*Answer can only contain numeric values

QUESTION: 25

For copper at 1000 K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (E_{F} = 7 eV)

Solution:

The probability F(E) of a state corresponding to energy E being occupied by an electron temperature T is given by

There fore

or

Thus

For copper, E_{F} = 700 eV

so that E = E_{F} + ΔE = 7.00 eV - 0.19 eV = 6.81 eV

*Answer can only contain numeric values

QUESTION: 26

The given figure shows a silicon transistor connected as a common emitter amplifier. The quiescent collector voltage of the circuit is approximately______V.

Solution:

*Answer can only contain numeric values

QUESTION: 27

A p-n junction diode in series with a 100 ohms resistor is forward biased so that a current of 100 mA flows. If the voltage across this combination is instantaneously reversed to 10 V at t = 0. the reverse current that flows through the diode at t = 0 is approximately________ mA.

Solution:

Reverse current at t = 0 when the voltage

is instantaneously reversed to - V_{R} = - 10 V is

Negative sign indicating reversal of current and voltage.

∴

*Answer can only contain numeric values

QUESTION: 28

The 6V zener diode shown in the figure, has zero zener resistance and a knee current of 5mA. The minimum value of R so that the voltage across it does not fall below 6 V i s _________ ohms.

Solution:

Output voltage is regulated to the zener voltage 6V.

Zener diode current, l_{ZK} = 5 mA

This is the minimum current drawn by zener.

Hence the load current l_{L} will be maximum.

∴ I_{L,max} = 80 - 5 = 75mA

∴

*Answer can only contain numeric values

QUESTION: 29

A voltage source V_{AB} = 4 sin ωt is applied to the terminal A and B of the circuit shown in the given figure. The diodes are assumed to be ideal. The impedance by the circuit across the terminal A and B is __________ kΩ.

Solution:

Diode D_{1} conducts through 10 kΩ on the extreme right and diode D_{2} is blocked. Diode D_{2}, conducts through the 10k Ω in the middle branch and diode D_{1} is blocked. Thus the source always sees a resistance of 10 kΩ.

*Answer can only contain numeric values

QUESTION: 30

A transistor with β = 45, is used with collector to base biasing with a quiescent value of 5 volt for V_{ce} . If V_{cc} = 24 volt, R_{c} = 10 kΩ and R_{e} = 270 ohm. The value of R_{b} is kΩ.

Solution:

Applying Kirchoff s low to collector-emitter circuit, we have

or 24 = (1 +45) l_{b} (10 - 10^{3} x 270)+ 5

or 19 = 46 l_{b} [10.27 x 10^{3}]

or

∴

From the figure

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