IIT JAM Physics MCQ Test 11 - Physics MCQ

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 = a³ = (0.28845)³ nm³
= 0.024 nm³ 
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 nm³

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

 

Explanation:

In a face-centered cubic (fcc) unit cell, there are 8 corners and 6 faces.

Each corner of the unit cell is shared by 8 adjacent unit cells, so each corner atom contributes 1/8 of an atom to a particular unit cell. Since there are 8 corners, this amounts to

8×(1/8) = 1 atom of type A.

Each face of the unit cell is shared by 2 unit cells, so each face-centered atom contributes 1/2 of an atom to a particular unit cell. Since there are 6 faces, this amounts to

6× (1/2) = 3 atoms of type B.

Therefore, the stoichiometry of the compound is AB₃, which corresponds to option 1.

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 d₁ = a and the distance of next nearest neighbour d₁ = a√2

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

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

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)

diode D₁ is always reverse biased 
Then V₀ -5V (for the complete cycle)

Voltage across R_L = 10V. 
R_L=5kΩ

In an n-type semiconductor, the minority hole concentration is given by:

N_D = Concentration of Donor impurity

In a p-type semiconductor, the minority electron concentration is given by:

N_A = Concentration of Acceptor impurity

Let x₁. x₂, x₃ 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 
X₁ : X_{2 ^:} X₃ = 3 : 2 : 6

Determine the output voltage in Fig.


= - (0.9) 10 = - 9V

Given, I_Z (min) = 5 mA
V₀ = 6 V = V_z
(according to given condition).
Maximum power dissipation means maximum zener current
i.e. I_z (max) = P_max V = 300 mW 6 V = 50 mA
from figure
I = V_in – V_z 50 = 9 – 6 50 = 60 mA
and I = IZ + IL
I_L (min) = I – I_z (max) = 60 – 50 = 10 mA
I_L (max) = I – I_z (min) = 60 – 5 = 55 mA
Hence alternative (C) is the correct choice.

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

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 I₁, and l₂, be the currents in resistors R₁, and R₂ and l₂ = 10 I\b

Now

and

Hence 

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.

Here α = 0.985

We know

∴ β = 66
Base current. 
Voltage across R₂ is given by

Voltage across R₁ is given by

Current flowing through R₁ and R₂


∴ Resistance, R₁ =  
Voltage across R_c = V_cc - V_ce - Ve
= 20 - 5 - 2 x 2 = 11 volt
Collector resistance, 

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 V₀ = 4 volt.

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 

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

0.720
Conductivity of an intrinsic semiconductor is given by 

or

= 2.5 x 10¹⁹ / m³
But

for C = 4.8 x 10²¹
and k_BT = 1.38 x 10^-23 x 300J = 
We have

or

and
E_g = 2 x 13.8 x 0.025875 = 0.720 eV.

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.

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

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.
∴ 

Diode D₁ conducts through 10 kΩ on the extreme right and diode D₂ is blocked. Diode D₂, conducts through the 10k Ω in the middle branch and diode D₁ is blocked. Thus the source always sees a resistance of 10 kΩ.

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

or 24 = (1 +45) l_b (10 - 10³ x 270)+ 5 
or 19 = 46 l_b [10.27 x 10³]
or 

∴ 
From the figure