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Extra Credit 23

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

    #58,  #64, #92

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

    #14, #37, #50


    Question #58

    The tungsten elements of incandescent light bulbs operate at 3200 K. At what wavelength does the filament radiate maximum energy?

    To solve for the wavelength of maximum energy, we need to use Wien's Displacement Law which relates temperature (\(T\)) to the wavelength (\(\lambda_{max} \)) with a proportionality constant called Wien's displacement constant (\(b = 2.89 \times 10^{-3}m  \hspace{1mm} K\)).

    \[\lambda_{max} = \dfrac{b}{T}\]

    Using \(T = 3200 K\), we can then solve for \(\lambda_{max}\):

    \[\lambda_{max} = \dfrac{2.89 \times 10^{-3}m  \hspace{1mm} K}{3200 K} \\ = 9.03 \times 10^{-7} m \\ = 903 nm\]

    The tungsten filament radiates maximum energy at 903 nm.

    Question #64

    What is the longest wavelength of radiation that can eject a photoelectron from potassium, given the work function of potassium 2.24 eV? Is it in the visible range?

    The work function (\(\phi\)) is the required energy to free an electron from an atom and it is equal to the threshold frequency (\(\nu_0\)) multiplied by Plank's constant (\(h = 6.626 \times 10^{-34} J \hspace{1mm} s\)). The threshold frequency is the lowest energy light particle needed to satisfy the work function.

    \[\phi = \frac{h}{\nu_0} \]

    We can change this equation to an expression with the wavelength by using the relationship that frequency is the inverse of the wavelength (\(\lambda\)) times the speed of light (\(c = 3.00 m \hspace{1mm} s^{-1}\)), \(\nu = \frac{c}{\lambda}\).

    \[\phi = \dfrac{h \hspace{1mm} c}{\lambda}\]

    Then we can rearrange this equation to solve for \(\lambda\).

    \[\lambda = \dfrac{h \hspace {1mm} c}{\phi}\]

    Before plugging in the values, the work function must be converted to units of Joules (\(1 eV = 1.60 \times 10^{-19} J\)).

    \[2.24 eV \times \dfrac{1.60 \times 10^{-19} J}{1 eV} = 3.594 \times 10^{-19} J\]

    Now, we can plug in the values and evaluate the equation.

    \[\lambda = \dfrac{(6.626 \times 10^{-34}J \hspace{1mm} s) (3.00 m \hspace{1mm} s^{-1})}{3.594 \times 10^{-19} J} \\ = 2.63 \times 10^{-7} m \\ = 263 nm\]

    The longest wavelength of radiation that can eject a photoelectron from potassium is 263 nm. It is not in the visible range because the visible range is from 380 nm to 700 nm. Instead, it is in the UV range.

    Question #92

    Determine the wavelength of the third Balmer line (transition from \(n = 5\) to \(n = 2\)).

    To solve for the wavelength, we must use the Rydberg Formula

    \[\tilde{\nu} = \dfrac{1}{\lambda} = R \left({\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}}\right) \quad n_1 = 1, 2, 3..., n_2 > n_1\]

    \tilde{\nu} is the wave number, which is just the inverse of the wavelength. R is the Rydberg constant (\(109,737 cm^{-1}\) or \(2.18 \times 10^{-18} J\)).

    Rearrange the Rydberg formula to solve for \(\lambda\).

    \[\dfrac{1}{\lambda} = R \left({\dfrac{n_2^2 - n_1^2}{n_2^2 n_1^2}}\right) \\ \lambda = \dfrac{1}{R} \left({\dfrac{n_2^2 \hspace{1mm} n_1^2}{n_2^2 - n_1^2}}\right)\]

    Now, we can plug in the values and evaluate the equation.

    \[\lambda = \dfrac{1}{109,737 cm^{-1}} \left({\dfrac{5^2 \hspace{1mm} 2^2}{5^2 - 2^2}}\right) \\ = \dfrac{1}{109,737 cm^{-1}} \left({\dfrac{25 \hspace{1mm} 4}{25 - 4}}\right) \\ = \dfrac{1}{109,737 cm^{-1}} \left({\dfrac{100}{21}}\right) \\ = 4.3394 \times 10^{-5} cm \\ = 433.94 nm\]

    The wavelength of the third Balmer line is 433.94 nm.


    Question #14

    Which aspects of the photoelectric effect cannot be explained by classical physics?

    Classical physics could not explain how the the kinetic energies of the electrons emitted were not affected by the intensity of the light but the frequency instead. The intensity affects the number of electrons emitted, which is unaffected by kinetic energy and frequency. It also could not explain how the photo electric effect could be observed immediately upon applying a light source.

    Question #37

    Do gravitational forces have a significant effect on atomic energy levels?

    No, gravitational forces have no significant effect on atomic energy levels. 

    Question #50

    Explain the importance of the Young double-slit experiment.

    The Young double-slit experiment proved that light was a wave.

    Extra Credit 23 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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