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Extra Credit 26

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

    #61,  #67, #95


    We first have the energy converted to Joules and equate the energy to planks constan and frequency

    \[E = 20keV = 2*10^{4} eV =h\nu \nonumber\]

    \[1eV = 1.60218*10^{-19} J\nonumber\]

    Then we calculate the frequency

    \[\nu = \frac{E}{h} = 4.836*10^{18} hz\nonumber\]

    Then we convert frequency to wavelength based on velocity being the speed of light

    \[\lambda = \frac{c}{\nu} = 6.20*10^{-11}m\nonumber\]


    We start with the equation for photoelectric effect

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

    The kinetic energy is 0 if we want to find the least amount of energy to dislodge an electron.

    \[0 = \frac{hc}{\lambda} - \phi \nonumber\]

    Plug in your wavelength and solve for the workfunction

    \[\phi = \frac{hc}{304*10^{-9}} \nonumber\]

    \[\phi = 6.53 * 10^{-19}J \nonumber\]



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

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

    We will also be using

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


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

    The longest wavelength

    The longest wavelength correlates with the least energy released which is the smallest jump of one energy state from the third excited state (n=4) to the next lowest excited state (n=3).

    Note: The jump can't be fractional because of quantization.

    \[n_{f} =3  \nonumber\]

    \[n_{i} =4  \nonumber\]

    \[\Delta E = -R_{H}(\frac{1}{3^2}-\frac{1}{4^2}) = -1.06 E-19 J \nonumber\]

    \[\Delta E =-1.05875 E-19 J \nonumber\]

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

    \[\lambda = 1.876E-6m\nonumber\]

    The shortest wavelength

    The longest wavelength correlates with the most energy released which is the largest jump from the third excited state (n=4) to the lowest excited state (n=1).

    Note: The jump can't be fractional because of quantization.

    \[n_{f} =1  \nonumber\]

    \[n_{i} =4  \nonumber\]

    \[\Delta E = -R_{H}(\frac{1}{3^2}-\frac{1}{1^2}) = -1.06 E-19 J \nonumber\]

    \[\Delta E =-2.041875 E-18 J \nonumber\]

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

    \[\lambda = 9.7E-8m\nonumber\]

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

    #17, #40, #53


    Discuss any similarities and differences between the photoelectric and the Compton effects.

    They are similar in that they both propose methods of how energy is transferred from photon to electron. The photoelectron effect is when an electron is able absorb the full energy of a photon and allows only allows for quantized packets of energy to be absorbed. The compton effect is slightly different where it allows the photon to act as more of a particle and transfers a portion of its energy then scatters.


    Speculate as to how the diffraction patterns of a typical crystal would be affected if gamma rays were used instead of X-rays.

    The diffraction pattern would be hard to analyze because the scattering would not scatter too much. The wavelength of gamma rays are so small that the angle of diffraction that results would not be enough to produce usable/readable scattering data.


    Do the photons of red light produce better resolution in a microscope than blue light photons? Explain.

    The resolution of blue light photons would be better because the wavelength would be shorter. As a result, the angle of diffraction of blue light would be smaller allowing for better resolution.

    Extra Credit 26 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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