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Chemistry LibreTexts

Solutions 14

  • Page ID
    47394
  • Unless otherwise specified, a “particle in a box” below refers to the ground-state (quantum number n = 1) in a box with walls at x=0 and x=a.  Justify your answers to questions with calculations.

    Q14.1

    1. For a particle in a box of size a, normalize the ground-state wavefunction
    2. Do the same for a particle in a box of size 2a

     

    a) \[\psi = Asin(\frac{n\pi}{L}x)\]

    \[\psi^2 = A^2\int_{0}^{a} sin^2(\frac{n\pi}{L}x) = 1\]

    \[A = \sqrt{\frac{2}{a}}\]

    b) \[\psi = Asin(\frac{n\pi}{L}x)\]

    \[\psi^2 = A^2\int_{0}^{2a} sin^2(\frac{n\pi}{L}x) = 1\]

    \[A = \sqrt{\frac{1}{a}}\]

    Q14.2

    1. Calculate the probability of finding the particle in the left half of the box, when the particle is in the ground-state.
    2. Without calculations (or with, if you prefer), is the probability of finding the particle (for the ground-state) in the interval a/4 < x < 3a/4 greater or less than one-half?

     

    a) \[\int_{0}^{L/2}(\frac{2}{L})Sin^2(\frac{n\pi}{L}x) = .5\] This can be seen without math from the fact that the probability density is symmetric so the left half is equal to the right half. 

    b) Greater than one-half since the probability density peaks at L/2 and goes to zero towards the walls of the box. 

    Q14.3

     What is the frequency of the photon that is absorbed by a particle in a box as it makes a transition from the n=2 to n=3 stationary state?  Estimate (or calculate) a numerical value for the case of an electron in a box 1-nm long.

    The energy of a particle in a box is:

    \[E = \frac{n^2h^2}{8mL^2}\]

    \[\deltaE = E_3 - E_2 = \frac{5h^2}{8mL^2} = \frac{33x10^{-68}J^2s^2}{(8)(9.1x10^{-31})(1x10^{-18})} = 4.5x10^{-20}J\]

    Q14.4

    For the n=100 stationary state, a) how many nodes does it have? Is the wavefunction symmetric (even function), antisymmetric (odd function) or neither about the midpoint of the box?

    a) 99 nodes

    b) Anti-symmetric since all even n wavefunctions are anti-symmetric. 

    Q14.5

    For a particle in a box (with infinite height walls) answer the following intuition based questions (in words, not equations):

    1. How many wavefunctions exist for this problem? An infinite number
    2. How many different energies exist for this problem? An infinite number
    3. Is there an specific energy for every possible wavefunction? Yes, each wavefunction has its own energy.
    4. Is there a specific wavefunction for every possible energy of the particle? Yes, for every possible energy of the particle there is a corresponding specific wavefunction
    5. How do the energies of the wavefunctions change when the box is made bigger? They get smaller.
    6. How do the energies of the wavefunctions change when the particle mass is made bigger? They get smaller.
    7. How do the difference in successive energies (e.g., \(E_{n=2}-E_{n=1}\)) with increasing mass of particle? These get smaller as well because they have the same form as the energies. 
    8. How do the difference in successive energies (e.g., \(E_{n=2}-E_{n=1}\)) with increasing box length of particle?These get smaller as well because they have the same form as the energies. 
    9. What is the probability of finding the particle in the box? 100%
    10. What is the probability of finding the particle outside the box? 0%
    11. How would the answers to the above two questions change if the box height were decreased (i.e., finite)? In a finite box there is some probability of being outside the box and less than 100% probability for being inside the box.

    Q14.6

    An electron has a kinetic energy of 12.0 eV. The electron is incident upon a rectangular barrier of height 20.0 eV and thickness 1.00 nm. What is the probability of the electron tunneling through the barrier? By what factor would the electron’s probability of tunneling through the barrier increase assuming that the electron absorbs all the energy of a 500-nm photon (Hint: remember conservation of energy).

    The probability of a particle tunneling through a barrier is:

    \[P = exp[-(\frac{4\pia}{h})(2m(V-E))^{1/2}]\]

    Where:

    \[V - E = 8 eV = 1.28x10^{-18} J\]

    \[m = 9.1x10^{-31} kg\]

    \[a = 1x10^{-9} m\]

    \[P = 2.54x10^{-13}\]