- 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:
- h\nu\left(n+\frac{3}{2}\right)
- 2h\nu\left(n+\frac{1}{2}\right)
- h\nu\left(n+\frac{1}{2}\right), n odd only
- h\nu\left(n+\frac{1}{2}\right), n even only
- 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
- The expectation value of the energy is \frac{5}{4}\hbar\omega
- Energy measurement of this state will always gives its value as \frac{5}{4}\hbar\omega
- The average value of energy is \hbar\omega
- The expectation value of energy is dependent on time
- 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:
- n=3, l=2
- n=2, l=1
- n=1, l=2
- n=2, l=2
- 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 :
- 30 eV
- 10 eV
- 4 eV
- \frac{1}{4} eV
- 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)
- Z(\mu,T)=e^{-\epsilon\beta}
- Z(\mu,T)=\left(1+e^{-\epsilon\beta}\right)^{-1}
- Z(\mu,T)=1+e^{-\epsilon\beta}
- Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}
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Friday, 27 January 2017
Problem set 62
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