JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year APPLIED PHYSICS Previous Question Paper (July/August - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
Explain a black body radiation with example.
Answer
A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It does not reflect or transmit any radiation. When in thermal equilibrium at a specific temperature, it emits radiation known as black body radiation, which depends solely on the temperature of the body and not on its shape, size, or material composition.
Key characteristics include:
- It is a perfect absorber and a perfect emitter of radiation.
- The spectral distribution of the emitted radiation follows Planck's Law, which describes the intensity of radiation as a function of wavelength and temperature.
- As the temperature of the black body increases, the peak of the emission spectrum shifts toward shorter wavelengths (Wien's Displacement Law) and the total energy emitted increases (Stefan-Boltzmann Law).
An excellent practical example of a black body is a Lummer-Pringsheim cavity, which consists of an enclosure with a very small hole. Any radiation entering the hole undergoes multiple internal reflections, making it highly unlikely to escape, thus effectively trapping the radiation. The radiation emerging from the hole is a close approximation of ideal black body radiation.
Derive an expression for Particle in one dimensional box.
Answer
Consider a particle of mass m confined to a one-dimensional box of length L, where the potential energy V(x) is zero inside the box (0 < x < L) and infinite outside. The time-independent Schrödinger wave equation is:
Inside the box (V=0), the equation simplifies to:
Let k² = 2mE/ħ². The general solution is ψ(x) = A sin(kx) + B cos(kx). Applying boundary conditions:
- At x = 0, ψ(0) = 0, which implies B = 0.
- At x = L, ψ(L) = 0, which implies A sin(kL) = 0. Since A ≠ 0, sin(kL) = 0, so kL = nπ, where n = 1, 2, 3...
Thus, k = nπ/L. Substituting k back into the energy expression:
The normalized wave function is found by setting the integral of |ψ|² over the box to 1:
Differentiate between Intrinsic and Extrinsic semiconductors.
Answer
Sign in to view the complete answer paper.
Write a note on Carrier generation and recombination.
Answer
Sign in to view the complete answer paper.
Explain the formation of Extrinsic semiconductors with diagram.
Answer
Sign in to view the complete answer paper.
Explain the formation Bipolar junction transistor with its characteristics.
Answer
Sign in to view the complete answer paper.
Describe the recombination mechanism in non-radiative semiconductors.
Answer
Sign in to view the complete answer paper.
Explain LED and its uses.
Answer
Sign in to view the complete answer paper.
Describe the recombination mechanism in radiative semiconductors.
Answer
Sign in to view the complete answer paper.
What is a "LASER"? What are the merits of a semiconductor laser?
Answer
Sign in to view the complete answer paper.
Explain the principle and working of He-Ne laser.
Answer
Sign in to view the complete answer paper.
Mention the various types of Losses in optical fibres.
Answer
Sign in to view the complete answer paper.
Explain permittivity and Dielectric constant for a magnetic material.
Answer
Sign in to view the complete answer paper.
Derive Clausius-Mossotti Equation.
Answer
Sign in to view the complete answer paper.
How many types of magnetic materials with examples- Classify?
Answer
Sign in to view the complete answer paper.
Mention eight applications of magnetic materials.
Answer
Sign in to view the complete answer paper.