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JNTU B.Tech I APPLIED PHYSICS Previous Question Paper (September/October - 2021 - R18)

B.Tech | I | I-I | R18 | September/October - 2021

Answer PaperPaper Code: 151AEMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Show that the energies of a particle in a one dimensional potential box are quantized.

10 marks

Answer

Consider a particle of mass m in a one-dimensional potential box of width a, where the potential V(x) = 0 for 0 ≤ x ≤ a and V(x) = ∞ elsewhere. The time-independent Schrödinger equation inside the box is:

d2ψ/dx2 + (2mE/ℏ2)ψ = 0

Let k2 = 2mE/ℏ2. The general solution is ψ(x) = A sin(kx) + B cos(kx). Applying boundary conditions ψ(0) = 0 and ψ(a) = 0:

  1. At x = 0, ψ(0) = B = 0. Thus, ψ(x) = A sin(kx).
  2. At x = a, ψ(a) = A sin(ka) = 0. Since A ≠ 0, we must have sin(ka) = 0, implying ka = nπ where n = 1, 2, 3, ...
  3. Substituting k = nπ/a into the expression for k2: (nπ/a)2 = 2mE/ℏ2.

Solving for E, we obtain the quantized energy levels:

En = (n2π2ℏ2) / (2ma2)

Since n must be an integer, the energy can only take discrete values, demonstrating that the energy of the particle is quantized.

b

Find the probability of finding a particle between 0.35a and 0.65a where ‘a’ is the width of the box and particle is in the first excited state.

5 marks

Answer

For the first excited state, n = 2. The normalized wavefunction is ψ2(x) = √(2/a) sin(2πx/a). The probability P is given by the integral of |ψ2(x)|2 from 0.35a to 0.65a:

P = ∫0.35a0.65a (2/a) sin2(2πx/a) dx

Using the identity sin2θ = (1 - cos 2θ)/2:

P = (1/a) ∫0.35a0.65a (1 - cos(4πx/a)) dx
P = (1/a) [x - (a/4π) sin(4πx/a)]0.35a0.65a

Evaluating at the limits:

P = (0.65 - 0.35) - (1/4π) [sin(2.6π) - sin(1.4π)]
P = 0.3 - (1/4π) [sin(0.6π) - sin(1.4π)]
P = 0.3 - (1/4π) [0.951 - (-0.951)] = 0.3 - (1.902 / 12.566) ≈ 0.3 - 0.1514 = 0.1486

The probability of finding the particle in the specified region is approximately 0.1486 or 14.86%.

215 marks
a

Describe the Davisson and Germer’s experimental for verification of matter waves.

10 marks

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b

Electrons are accelerated through 344 volts and are reflected from a crystal. The first reflection maximum occurs when glancing angle is 60°. Determine the spacing of the crystal. Given h = 6.62 × 10-34 Joule-sec, e = 1.6 × 10-19 C and me = 9 × 10-31 kg.

5 marks

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315 marks
a

What is a PN-junction diode? Discuss the V-I characteristics of a diode in both the biasing conditions.

10 marks

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b

Explain advantage of Zener diode over P-N junction diode.

5 marks

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415 marks
a

Explain the Hall effect in metal? Derive the formulae to determine Hall coefficient and mobility of electrons.

10 marks

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b

An n-type germanium sample has a donor density of 1021/m3. It is arranged in a Hall experiment having magnetic field of 0.5T and the current density is 500 A/m2. Find the Hall voltage if the sample is 3mm wide.

5 marks

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515 marks
a

What is an LED? Explain the working of LED with a neat diagram.

10 marks

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b

Write a short note on solar cell.

5 marks

Answer

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615 marks
a

What are semiconductor diode lasers? Describe the construction and working of a semiconductor laser with energy band diagram.

10 marks

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b

Discuss advantages of diode lasers over gas lasers.

5 marks

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715 marks
a

Draw the block diagram of an optical fibre communication system and explain the function of each block.

10 marks

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b

Consider a fibre having a core of index 1.48, a cladding of index 1.46 and has a core diameter of 30 µm. Show that all rays making an angle less than 9.43° with the axis will propagate through the fibre.

5 marks

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815 marks
a

State and explain the basic laws of electromagnetism in their integral form.

10 marks

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b

Distinguish between conduction current and displacement current.

5 marks

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This page contains the solved question paper with answers of APPLIED PHYSICS from JNTU for B.Tech ITE I I-I, September/October - 2021, R18.

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