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Find the average energy per free electron (Etot/Nd), as a fraction of the

Fermi energy. Answer:(3/5)EF

Short Answer

Expert verified

The average energy per free electron isEtot/NqEF=35EF

Step by step solution

01

Formula used

Find ratio of average energy per free electron and Fermi energy.

We know:

Etot=h232Nq5/3102mV-2/3. 鈥(i)

EF=h22m3p22/3 鈥(颈颈)

02

Finding the average energy per free electron

Rearranging the terms in the equation 1:

Etot=h2V22m0kFK4dk=h2kF5V102mV-2/3=h32Nd5/3102mV-2/3

Density is given by:

=NdV

Now, average energy per free electron is:

Etot/NqEF=h232Nq5/3102m1Nq2m32Nq/V2/3=35EF

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Most popular questions from this chapter

(a) If aandb are orthogonal, and both normalized, what is the constant A in Equation 5.10?

(b) Ifrole="math" localid="1658225858808" a=b (and it is normalized), what is A ? (This case, of course, occurs only for bosons.)

The density of copper is8.96g/cm3,and its atomic weight is63.5g/mole

(a) Calculate the Fermi energy for copper (Equation 5.43). Assume d = 1, and give your answer in electron volts.

EF=22m3蚁蟺22/3 (5.43).

(b) What is the corresponding electron velocity? Hint: SetEF=1/2mv2Is it safe to assume the electrons in copper are nonrelativistic?

(c) At what temperature would the characteristic thermal energyrole="math" localid="1656065555994" (kBT,wherekBkBis the Boltzmann constant and T is the Kelvin temperature) equal the Fermi energy, for copper? Comment: This is called the Fermi temperature,TF

. As long as the actual temperature is substantially below the Fermi temperature, the material can be regarded as 鈥渃old,鈥 with most of the electrons in the lowest accessible state. Since the melting point of copper is 1356 K, solid copper is always cold.

(d) Calculate the degeneracy pressure (Equation 5.46) of copper, in the electron gas model.

P=23EtotV=232kF5102m=322/325m5/3

Suppose you had three particles, one in statea(x), one in stateb(x), and one in statec(x). Assuming a,b, andc are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row isa(x1),b(x1),c(x1) , etc., whese second row isa(x2),b(x2),c(x2) , etc., and so on (this device works for any number of particles).

(a) Show that for bosons the chemical potential must always be less than the minimum allowed energy. Hint:n(o)cannot be negative.

(b) In particular, for the ideal bose gas, (T)<0for allT. Show that in this case(T)monotonically increases asTdecreases, assumingNandVare held constant.

Hint: Study Equation5.108, with the minus sign.


(c) A crisis (called Bose condensation) occurs when (as we lowerT )role="math" localid="1658554129271" (T)hits zero. Evaluate the integral, for=0, and obtain the formula for the critical temperatureTc at which this happens. Below the critical temperature, the particles crowd into the ground state, and the calculational device of replacing the discrete sum (Equation5.78) by a continuous integral (Equation5.108) losesits validity 29.

Hint:role="math" localid="1658554448116" 0xs-1ex-1dx=(s)(s)
where 螕 is Euler's gamma function and is the Riemann zeta function. Look up the appropriate numerical values.


(d) Find the critical temperature for 4He. Its density, at this temperature, is 0.15 gm / cm3. Comment: The experimental value of the critical temperature in 4He is 2.17 K. The remarkable properties of 4He in the neighborhood of Tc are discussed in the reference cited in footnote 29.

(a) Calculate<1/r1-r2>for the state0(Equation 5.30). Hint: Dod3r2integral

first, using spherical coordinates, and setting the polar axis alongr1, so

that

0r1,r2=100r1100r2=8蟺补3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2肠辞蝉胃2.

The2integral is easy, but be careful to take the positive root. You鈥檒l have to

break ther2integral into two pieces, one ranging from 0 tor1,the other fromr1to

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it toE0(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we鈥檙e still working with an approximate wave function, so don鈥檛 expect perfect agreement.)

E0=8-13.6eV=-109eV(5.31).

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