ExamVeda
Login
Home

For a Fermi gas of N particles in three-dimensions at T...

This question belongs to Engineering Physics Thermodynamics And Statistical Physics
Thermodynamics and Statistical Physics
?

For a Fermi gas of N particles in three-dimensions at T = 0 K, the Fermi energy EF is proportional to

Answer & Solution
Correct Answer: Option A
Let's discuss the solution. Join the discussion
Examveda
Question posted by Examveda
Community

Join the Discussion

No comments yet Be the first to discuss this question.
Keep Learning

Related Questions on Thermodynamics and Statistical Physics

A system has two energy levels with energies E and 2E. The lower level is four-fold degenerate while the upper level is doubly degenerate. If there are N non-interacting classical particles in the system, which is in thermodynamic equilibrium at temperature T, the fraction of particles in the upper level is
A. $$\frac{1}{{1 + {e^{ - \varepsilon /{k_B}T}}}}$$ B. $$\frac{1}{{1 + 2{e^{\varepsilon /{k_B}T}}}}$$ C. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} + 4{e^{2\varepsilon /{k_B}T}}}}$$ D. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} - 4{e^{2\varepsilon /{k_B}T}}}}$$
A system of non-interacting Fermi particles with Fermi energy EF has the density of states proportional to √E, where E is the energy of a particle. The average energy per particle at temperature T = 0 is
A. $$\frac{1}{6}{E_F}$$ B. $$\frac{1}{5}{E_F}$$ C. $$\frac{2}{5}{E_F}$$ D. $$\frac{3}{5}{E_F}$$
A piston containing an ideal gas is originally in the state X (see figure). The gas is taken through a thermal cycle X → Y → X as shown
Thermodynamics and Statistical Physics mcq question image
The work done by the gas is positive, if the direction of the thermal cycle is
A. clockwise B. counter-clockwise C. neither clockwise nor counter-clockwise D. clockwise from X → Y and counter-clockwise from Y → X
A photon gas is at thermal equilibrium at temperature T. The mean number of photons in an energy state $$\varepsilon = h\omega $$  is
A. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1$$ B. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1$$ C. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1} \right]^{ - 1}}$$ D. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1} \right]^{ - 1}}$$
More Related Questions on Thermodynamics and Statistical Physics