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Physics Resonance: Problem set 62 -->

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Friday, 27 January 2017

Problem set 62

  1. The energy levels of one-dimensional harmonic oscillator with potential V(x)=\frac{1}{2}kx^2 are given by h\nu\left(n+\frac{1}{2}\right) with n=0,1,2,\dots. If the potential is changed to V(x)=\infty for x<0 and V(x)=\frac{1}{2}kx^2 for x>0, the energy levels now, will be given by:
    1. h\nu\left(n+\frac{3}{2}\right)
    2. 2h\nu\left(n+\frac{1}{2}\right)
    3. h\nu\left(n+\frac{1}{2}\right), n odd only
    4. h\nu\left(n+\frac{1}{2}\right), n even only
  2. At t=0 a one-dimensional harmonic oscillator is in a state given by:\Psi(x,0)=\frac{1}{2}u_0(x)+i\frac{\sqrt{3}}{2}u_1(x) where u_0 and u_1 are first two normalized eigen states. (\omega is natural angular frequency of the oscillator). Then
    1. The expectation value of the energy is \frac{5}{4}\hbar\omega
    2. Energy measurement of this state will always gives its value as \frac{5}{4}\hbar\omega
    3. The average value of energy is \hbar\omega
    4. The expectation value of energy is dependent on time
  3. If the \phi dependent part of the eigen function of an electron in a Hydrogen atom is e^{2i\phi}, then the minimum principal and minimum orbital angular momentum quantum numbers n and l respectively for this eigen function will be:
    1. n=3, l=2
    2. n=2, l=1
    3. n=1, l=2
    4. n=2, l=2
  4. The ground state energy of a particle in an infinite square well is 1eV. If four particles obeying Bose-Einstein statistics are kept in this well, then the ground state energy will be :
    1. 30 eV
    2. 10 eV
    3. 4 eV
    4. \frac{1}{4} eV
  5. Consider a system in contact with a heat and particle reservoir. It may be unoccupied or occupied by one particle with energy 0 and \epsilon. The grand partition function will be (\beta=1/kT)
    1. Z(\mu,T)=e^{-\epsilon\beta}
    2. Z(\mu,T)=\left(1+e^{-\epsilon\beta}\right)^{-1}
    3. Z(\mu,T)=1+e^{-\epsilon\beta}
    4. Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}

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