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No. of spinless particles in the left half of a box

  1. Nov 9, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-11-9_17-33-44.png
    How to solve question no. 35?
    2. Relevant equations


    3. The attempt at a solution
    Since the particle is spinless, spin = 0 , this means that the particle is a boson.
    Applying Bose - Einstein distribution function,
    ## f(E) = \frac1 { e^{\beta ( E - \mu)} -1}##
    I can get the value of ##\beta ## and ##\mu## [as this distribution function tells us the no. of particles having the energy E of a given system.
    Since I have LHS for two given values of E, I have 2 eqns. and hence I can determine ##\beta ## and ##\mu## .
    Now, what to do?
    How to connect this with the given information in question?
     
  2. jcsd
  3. Nov 9, 2017 #2

    DrClaude

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    Staff: Mentor

    You have a system with a fixed energy, not a fixed temperature, hence to BE distribution does not apply. Start by considering the eigenstates of a particle in a box.
     
  4. Nov 9, 2017 #3
    The eigen energy of a particle in a box is
    ## E_n = \frac{ n^2 h^2}{2 m a^2} = n^2 \epsilon_0##
    This gives that the particles are in the states n =2,15.
    Do I have to calculate the probability of getting a particle with energy ## E_2 ~ and~ E_{15}## each in the left half region and then I will multiply the probabililty with 1000 and add the two no.?
     
  5. Nov 9, 2017 #4

    DrClaude

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    Staff: Mentor

    Basically yes, assuming that the observation time is chosen randomly (or consider the result "on average").
     
  6. Nov 9, 2017 #5
    I didn't get the assumption. Why do we need the assuption?
     
  7. Nov 9, 2017 #6

    DrClaude

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    Staff: Mentor

    Because a superposition of stationary states of different energies is not itself a stationary state. The particles will be sloshing from left to right.
     
  8. Dec 5, 2017 #7
    i calculated the probability of finding a particle on the left half of the box, and it is always 1/2 for a stationary state, for all n..
    so, for the particles with energy 4e, the no of particles on the left half 100/2 = 50, for 225e, 900/2 = 450, total 500..

    best of luck for the TIFR GS btw..
     
  9. Dec 5, 2017 #8
    Thanks.
     
  10. Dec 5, 2017 #9
    Is the probability of finding the particle between ## x_i and x_f ## for 1 D box ##\frac { x_f - x_i} { L} ##, where L is the length of the box?

    I am not caclculating it. I just want to see it this way if it is so.
     
  11. Dec 6, 2017 #10
    I don't know the closed way to represent the probability, what I did was integration of psi* multiplied by psi with the limit - L to 0,where psi is the normalized wavefunction of 1d potential box.. 2L is the length of the box..
     
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