Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 10, Problem 12P
(a)
To determine
The average energy per photons in thermal equilibrium.
(b)
To determine
The average photon energy in electron volt.
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Consider photons at temperature T = 300K in a cubic box of volume 1 m' with periodic boundary conditions.
a) Find the total number of photons in the lowest orbital state. What is the total energy of these photons?
Hint: The 1-particle energy of photons is ɛ(k,s)=ħck =
hc
, independent of polarization s.
Consider the Bose-Einstein distribution function (with u= 0) for the lowest-energy orbital states
2л
k, = (1,0,0), k, =(0,1,0), k, =(0,0,1). Find the total number of photons that occupy
L
L
L
these states, taking into account that each of the orbital states has 2 polarizations s.
b) Find the number of photons in a single orbital state with wavelength 2 = 5000 Å.
What is the total energy of these photons?
We can approximate an electron moving in a nanowire (a small, thin wire) as a one-dimensional infi nite square-well potential. Let the wire be 2.0 μm long. The nanowire is cooled to a temperature of 13 K, and we assume the electron’s average kinetic energy is that of gas molecules at this temperature ( 3kT/2). (a) What are the three lowest possible energy levels of the electrons? (b) What is the approximate quantum number of electrons moving in the wire?
Consider 1.50 mols of a substance in thermal equilibrium at a temperature
of 87.0 °C whose atoms can occupy only two energy levels separated by 5.00x102 eV,
where 1 eV (electron volt) is an energy unit equal to 1.60x10 19 J. kB = 1.38x10 23 J/K ,
%3D
NA 6.02x1023 mol1
Eg
a) Find the probability of occupation of the ground state
with energy E1 and the 1st excited state with energy E2?
b) How many atoms in the substance (on average) are in
5.00x10 eV
the ground state and in the 1st excited state?
E
Chapter 10 Solutions
Modern Physics
Ch. 10 - Prob. 1QCh. 10 - Prob. 2QCh. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3PCh. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 8PCh. 10 - The light from a heated atomic gas is shifted in...
Ch. 10 - Prob. 10PCh. 10 - To obtain a more clearly defined picture of the...Ch. 10 - Prob. 12PCh. 10 - Prob. 14PCh. 10 - Prob. 15PCh. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 26P
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- A single electron of mass m can move freely along a one-dimensionl gold nanowire. Let x be the position coordinate of the electron along the wire. (a) Let ø (x) be the wave function of the electron. The quantity |ø (x)| has dimensions of inverse length. Explain very briefly the meaning of this quantity as a probability density. (b) Let us assume that $ (x) = A sin (3kox) (2) where A and ko are fixed, positive constants. Establish whether this wave function represents an eigenstate of momentum p. Justify your answer. Hint: the momentum operator is p -ih. - (c) Establish whether the wave function (x) given in Eq. (2) represents an eigenstate of kinetic energy K. Justify your answer. Hint: the kinetic energy operator is K = p²/2m. (d) Let us now assume that the gold nanowire mentioned above is not infinite, but extends over a finite length from r= 0 to x = L. Inside this region, the potential energy of the electron is zero, but outside this region the potential energy is infinite…arrow_forwardAn atom of iron has a radius of 156. pm and the average orbital speed of the electrons in it is about ×5.7*10^7 m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of iron. Write your answer as a percentage of the average speed, and round it to 2 significant digits.arrow_forwardImpurities in solids can be sometimes described by a particle-in-a-box model. Suppose He is substituted for Xe, and assume a particle-in-a-cubic-box model, the length of whose sides is equal to the atomic diameter of Xe (≈ 2.62 Å). Compute the lowest excitation energy for the He atom’s motion. (This is the energy difference between the ground state and the first excited state.)arrow_forward
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