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Extra Credit 22

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

    #57,  #63, #91



    We use Wein's law to solve this problem

     \[ I = \frac{2 \pi hc^2}{\lambda^5} * \frac{1}{e^{\frac{hc}{\lambda k * T}}-1}\nonumber\]

    To find the peak, we take the derivative of the intensity with respect to \[\lambda\nonumber\]. We know that at maximums, the derivative is 0.

    We let

    \[a= 2\pi hc^2 \nonumber\]

    \[b= \frac{hc}{kT} \nonumber\]

    then we have

    \[0 = \frac{be^\frac{b}{\lambda}}{\lambda(e^\frac{b}{\lambda}-1)}-5 \nonumber\]

    We plug in 0.4 micrometers for lambda and solve or T

    We get

    \[T = 7245 K \nonumber\]



    For be we use the same equation

    \[0 = \frac{be^\frac{b}{\lambda}}{\lambda(e^\frac{b}{\lambda}-1)}-5 \nonumber\]

    and plug in T as 800K to find lambda. We get

    \[\lambda = 3.62\mu m \nonumber\]



    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



    Use the equation for all 3 parts

    \[\frac{1}{\lambda}= -R(\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}) \nonumber\]

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

    \[\Delta E = -R*hc* (\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}) \nonumber\]

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

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


    \[n_{f} =4  \nonumber\]

    \[n_{i} =3  \nonumber\]

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

    1.06 E-19 J released


    \[n_{f} =1  \nonumber\]

    \[n_{i} =2  \nonumber\]

    \[\Delta E = -R_{H} (\frac{1}{1^2}-\frac{1}{2^2}) = -1.63E-18 J \nonumber\]

    -1.63E-18 J released


    \[n_{f} =\infty  \nonumber\]

    \[n_{i} = 3 \nonumber\]

    \[\Delta E = -R_{H} (\frac{1}{\infty^2}-\frac{1}{3^2}) =  2.42E-19 J\nonumber\]

    2.42E-19 J absorbed

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

    #13, #36, #49


    Increasing the temperature of a photoelectrode would decrease the workfunction of the metal. This would lead to the threshold frequency at which light can still knock off an electron to decrease.


    Visible light does not have the right frequency range to be absorbed by glass, so the photons pass through the glass. Wood, on the other hand, are able to absorb visible light frequencies.


    The particle nature of radiation is supported by an X-ray scattering experiment where energy loss was observed by the observed decrease in frequency and was dependent on the angle of collision. By the Compton effect, we conclude that the radiation acted as a particle in that instance.

    Extra Credit 22 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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