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Find the energy at the bottom of the first allowed band, for the case=10 , correct to three significant digits. For the sake of argument, assume a=1eV.

Short Answer

Expert verified

The energy at the bottom of the first band is 0.345eV.

Step by step solution

01

Define the Schrödinger equation

  • A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.
  • The time-dependent Schrodinger equation is represented as

Iddt|t>=H^|(t)>

02

Calculating the minimum energy of first band

The minimum energy of the first band required z,f(z)=1

And f(z) is given by

fz=cosz+sinzzz=ka,z蟺尾=10=尘伪补h2fz=cosz+10sinzz

03

Using MATLAB to calculate the minimum energy of first band

UsingMATLABwegetz=2.6276.EnergyisgivenwithE=h2k22m=h22m(za)2 =z22ah2尘尾=z22aa=1eVE=2.627672201eVE=0.345eV

Therefore the energy at the bottom of the first band is 0.345eV.

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Most popular questions from this chapter

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).

Imagine two non interacting particles, each of mass , in the one dimensional harmonic oscillator potential (Equation 2.43). If one is in the ground state, and the other is in the first excited state, calculate (x1-x2)2assuming
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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.

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(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.

Discuss (qualitatively) the energy level scheme for helium if (a) electrons were identical bosons, and (b) if electrons were distinguishable particles (but with the same mass and charge). Pretend these 鈥渆lectrons鈥 still have spin 1/2, so the spin configurations are the singlet and the triplet.

(a) Suppose you put both electrons in a helium atom into the n=2state;

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