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Extra Credit 28

  • Page ID
    195754
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     https://phys.libretexts.org/Bookshel...ves_(Exercise)

    Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

    #63,  #96, #107

    #63

    We start with the equation

    \[KE = h\nu - \phi \nonumber\]

    Since we want the largest wavelength, we assume that the KE is the smallest possible with is 0 in this case giving us

    \[h\nu = \phi \nonumber\]

    Frequency is speed of light divided by wavelength so we substitute frequency with wavelength.

    \[h* (\frac{c}{\lambda}) = \phi \nonumber\]

    Given at the binding energy of silver is 4.73 eV converted to 7.578295e-19 Joules.We solve for lambda to get

    \[\lambda =262 nm  \nonumber\]

    This is outside the range for visible light which is 380 to 700 nm

    #96

    \[\Delta E = -R_{H} (\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}) \nonumber\]

    where RH is 2.178E-18 J and the n are the final and initial energy states respectively

    We will also be using

    \[\Delta E = \frac{hc}{\lambda}\nonumber\]

    so

    \[\lambda = \frac{hc}{\Delta E}\nonumber\]

     

    \[n_{f} = \infty  \nonumber\]

    \[n_{i} = 1  \nonumber\]

    \[\Delta E = -R_{H}(\frac{1}{\infty^2}-\frac{1}{1^2}) = -1.06 E-19 J \nonumber\]

    \[\Delta E = 2.178 E-18 J \nonumber\]

    \[\lambda = \frac{hc}{\Delta E}\nonumber\]

    \[\lambda = 9.10E-8m\nonumber\]

    #107

    We use the formula for de Broglie wavelength

    \[\lambda = \frac{h}{p} = \frac{h}{mv} \nonumber\]

    plug in the mass of an electron for m an dtehh speed for v where h is the plank constant

    \[\lambda = \frac{h}{(9.10938356*10^{-3} )( 5.0*10^{6})} \nonumber\]

    \[\lambda = 1.4556*10^{-10}m \nonumber\]

    Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

     https://phys.libretexts.org/Bookshel...ves_(Exercise)

    #64,  #97, #108

    #64

    We start with the equation

    \[KE = h\nu - \phi \nonumber\]

    Since we want the largest wavelength, we assume that the KE is the smallest possible with is 0 in this case giving us

    \[h\nu = \phi \nonumber\]

    Frequency is speed of light divided by wavelength so we substitute frequency with wavelength.

    \[h* (\frac{c}{\lambda}) = \phi \nonumber\]

    Given at the binding energy of potassium is 2.24 eV converted to 3.58888e-19 Joules.We solve for lambda to get

    \[\lambda =5.54 *10^{-7} m  \nonumber\]

     

    #97

    For an electron in a hydrogen atom in the n=2 state, compute:

    (a) the angular momentum;

    L = 2E-33 Js

    (b) the kinetic energy;

      3.4 eV

    (c) the potential energy; and

    -6.8 eV

    (d) the total energy

    -3.4 eV

    #108

    What is the de Broglie wavelength of an electron that is accelerated from rest through a potential difference of 20 kV?

    0.274 nm

     


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