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Extra Credit 14

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

    #51, #64, #81

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


    Problem 51

    What is the ground state energy (in eV) of an αα-particle confined to a one-dimensional box the size of the uranium nucleus that has a radius of approximately 15.0 fm?

    The expression for ground state energy in a one dimensional box is as follows: 

    \[ E_0 =\pi^2\hbar^2/2mL^2\]


    The constants involved in this are going to be as follows:


    \[\hbar = 1.0546\times10^-34  J/s\]

    \[m = 6.64\times10^-27 kg\]

    \[L = 30\times10^-15  m\]


    Plugging into the equation yields: 

    \[E_{0}=\dfrac{\pi ^{2}\left( 1.0546\times 10^{-34}\dfrac{J}{s}\right) ^{2}}{\left( 2\right) \left( 6.64\times 10^{-27}kg\right) \left( 30\times 10^{-15}m\right) ^{2}}\]

    \[=9.184\times 10^{-15}J\]



    Problem 64

     Find the expectation value ⟨x^2⟩ of the square of the position for a quantum harmonic oscillator in the ground state.


    The expression for the expectation value of <x^2> is given as follows:

    \[\langle x^{2}\rangle =\langle \psi_0\left| x^{2}\right| \psi_0\rangle\]

    The normalized wave function for the quantum harmonic oscillator is as follows:

    \[\psi _0\left( x\right) =\left( \dfrac{m\omega }{\pi \hbar }\right) ^{\dfrac{1}{4}}e^{-\dfrac{m\omega }{2\hbar }x^{2}}\]


    Given this to find the expectation value we will need to evaluate this integral:

    \[\langle x^{2}\rangle =\int ^{\infty }_{-\infty }\psi_0^{\ast }\left( x\right) x^{2}\psi _{0}\left( x\right) dx\]


    Plugging in the appropriate wave functions gives us this:

    \[=\sqrt{\dfrac{m\omega }{\pi \hbar }}\int ^{\infty }_{-\infty }x^{2}e^{-\dfrac{m\omega }{\hbar }x^{2}}dx\]

    Then doing the integral yields this and multiply by the scalar to get the answer:

    \[\int ^{\infty }_{-\infty }x^{2}e^{-\dfrac{m\omega x^{2}}{\hbar }}dx=\dfrac{\sqrt{\pi }}{2\left( \dfrac{m\omega }{\hbar }\right) ^{\dfrac{3}{2}}}\]

    \[\langle x^{2}\rangle =\sqrt{\dfrac{m\omega}{\pi \hbar }}\left( \dfrac{\sqrt{\pi }}{2\left( \dfrac{m\omega }{\overline{n}}\right) ^{\dfrac{3}{2}}}\right) =\dfrac{\hbar }{2m\omega }\]


    Problem 81

    Part A

    We can use a 1 dimensional box approximation again here:

    \[ E_0 =\pi^2\hbar^2/2mL^2\]

    Our constants from problem 51 all remain the same except m and L which are now:

    \[m = 0.20 kg\]

    \[L = 1.5 m\]

    Plugging this all into our equation we get:

    \[E_{0}=\dfrac{\pi ^{2}\left( 1.0546\times 10^{-34}\dfrac{J}{s}\right) ^{2}}{\left( 2\right) \left( 0.20 kg\right) \left( 1.5 m\right) ^{2}}\]

    \[=1.22\times 10^{-67}J\]

    The energy equation then can be rearranged for time into:


    Plugging into the equation with our previous energy value

    \[=\dfrac{1.5m}{\sqrt{2\left( \dfrac{1.22\times 10^{-67}J}{0.20kg}\right) }}\]

    \[=1.37\times 10^{33}s\]

    \[\left( 1.37\times 10^{33}s\right) \left( \dfrac{1\min }{605}\right) \left( \dfrac{1hr}{60\min }\right) \left( \dfrac{1days}{24hr}\right) \left( \dfrac{1year}{365days}\right)\]

    \[=4.34\times 10^{25}years\]

    The age of the universe is 13.7 billion years old so the orders of magnitude for the answer are 15 orders of magnitude over. This is a very large difference!

    Part B

    The transition of energy will be given by the following equation

    \[\Delta E=\left( n_{2}^{2}-n_{1}^{2}\right) \left( \dfrac{\pi \hbar ^{2}}{2mL^{2}}\right)\]

    We assumed our initial n value was 1 for part A, for part B, the second n value will be 2. This gives us this equation:

    \[\Delta E=\dfrac{\pi ^{2}\hbar ^{2}}{2mL^{2}}\left( 2^{2}-1^{2}\right)\]

    Solving for the transition energy gives us this:

    \[=\dfrac{3\pi ^{2}\hbar ^{2}}{2mL^{2}}\]


    \[=3.6\times 10^{-67}J\]

    The KE of a ball moving at 2 m/s is given:

    \[\begin{aligned}KE=\dfrac{1}{2}mv^{2}\ =\dfrac{1}{2}\left( 0.2kg\right) \left( 2 m/s\right) ^{2}\ =.4 J\end{aligned}\]

    It is many orders of magnitude higher than quantizing the ball. Up to 67 orders of magnitude!



    #5, #13, #28

    Problem 5

    Two cavity radiators are constructed with walls made of different metals. At the same temperature, how would their radiation spectra differ?


    By Weins Displacement law, the radiation spectra depends only upon temperature. The substance does not affect it.

    Problem 13

    Speculate how increasing the temperature of a photoelectrode affects the outcomes of the photoelectric effect experiment.


    As temperature increases, so does the the frequency, so the ejected electrons from a photoelectrode would be greater and they would have greater Kinetic Energy.

    Problem 28

     Hydrogen accounts for about 75% by mass of the matter at the surfaces of most stars. However, the absorption lines of hydrogen are strongest (of highest intensity) in the spectra of stars with a surface temperature of about 9000 K. They are weaker in the sun spectrum and are essentially nonexistent in very hot (temperatures above 25,000 K) or rather cool (temperatures below 3500 K) stars. Speculate as to why surface temperature affects the hydrogen absorption lines that we observe.


    In very hot stars, the hydrogen atoms are no longer in their gas phase and have reached an even greater thermal phase called plasma, which causes their electrons to no longer be linked to a specific nucleus, and therefore the probability of observing spectra is nonexistent, because the hydrogen atoms have been ionized in the sense that they no longer have electrons to pull down to their ground state. There is no longer an electron that can transition into a lower energy state and emit a photon. For colder stars, the electrons do not have as much energy as their 9000 K counterpart therefore, the necessary energy needed to see an energy transition and emission of the photon.

    Extra Credit 14 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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