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Extra Credit 17

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

    #54, #67, #84

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

    Question 54:

    If the energy of the first excited state of the electron in the box is 25.0 eV, what is the width of the box?


    To solve this problem, we will calculate the electron's energy in terms of the width and set it equal to the provided energy.

    The nth energy level of the quantum PIB is given by \begin{equation} E_{n}=\dfrac{\hbar^2\pi^2}{2mL^2}\, n^2=\dfrac{h^2}{8mL^2}n^2 \end{equation}

    (where n = 1, 2, 3, etc.)


    Subsituting the electron mass into the first excited state (n = 2):

    \begin{align*}E_{2} &=\dfrac{h^2}{8m_{e}L^2}(2)^2 \\\\&=\dfrac{(4)(6.626*10^{-34}\ m^2kg/s)^2}{(8)(9.11*10^{-31}\ kg)L^2}\\\\&=\dfrac{2.41*10^{-37}\ J*m^2}{L^2}\end{align*}

    Setting equal to the target energy value:

    \begin{equation*}\dfrac{2.41*10^{-37}\ J*m^2}{L^2}=25.0\ eV*\dfrac{1.602*10^{-19}\ J}{1\ eV}=4.01*10^{-18}\ J \end{equation*}

    Solving for L:

    \begin{align*}L^2 &= \dfrac{2.41*10^{-37}\ J*m^2}{4.01*10^{-18}\ J} \\\\L^2 &= 6.02*10^{-20}\ m^2 \\\\L &=2.453 *10^{-10}\ m \end{align*}

    The width of the box is .245 nm

    Question 67:

    Estimate the ground state energy of the quantum harmonic oscillator by Heisenberg’s uncertainty principle. Start by assuming that the product of the uncertainties Δx and Δp is at its minimum. Write Δp in terms of Δx and assume that for the ground state x≈Δx and p≈Δp, then write the ground state energy in terms of x. Finally, find the value of x that minimizes the energy and find the minimum of the energy.

    Recall Heisenberg's Uncertainty Principle:

    \begin{equation}\Delta x \Delta p >= \dfrac{\hbar}{2}\end{equation}

    Assuming that the product of the uncertainties is at its minimum:

    \begin{align*}\Delta x \Delta p &= \dfrac{\hbar}{2}\\\\ \Delta p &= \dfrac{\hbar}{2\Delta x}\end{align*}


    Approximating x≈Δx and p≈Δp:

    \begin{align*}x &≈ \Delta x \\ p &≈ \dfrac{\hbar}{2x}\end{align*} 


    Total energy is given by:

    \begin{equation} E = KE + PE \end{equation}

    For a harmonic oscillator:

    \begin{equation} PE = \dfrac{1}{2}kx^2 \end{equation}

    Putting it all together:

    \begin{align*}E &= \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2\\\\&= \dfrac{p^2}{2m} +\dfrac{1}{2}kx^2\\\\&= \dfrac{\hbar^2}{8mx^2} + \dfrac{1}{2}kx^2 \end{align*}

    To minimize the energy, differentiate with respect to x:

    \begin{align*} \dfrac{dE}{dx} &= \dfrac{d}{dx}(\dfrac{\hbar^2}{8mx^2} + \dfrac{1}{2}kx^2)\\\\&= \dfrac{-\hbar^2}{4mx^3} + kx\end{align*}

    And set equal to 0:

    \begin{align*}\dfrac{-\hbar^2}{4mx^3} + kx = 0 \\\\kx = \dfrac{\hbar^2}{4mx^3}\\\\4mkx^4 = \hbar^2\\\\x=(\dfrac{\hbar^2}{4mk})^{1/4}\end{align*}

    The energy is minimized at this value of x, and is equal to:

    \begin{align*}E &= \dfrac{\hbar^2}{8m(\dfrac{\hbar^2}{4mk})^{1/2}}+\dfrac{1}{2}k(\dfrac{\hbar^2}{4mk})^{1/2}\\\\&= \dfrac{2\hbar^2\sqrt{mk}}{8m\hbar} + \dfrac{k\hbar}{4\sqrt{mk}}\\\\&= \dfrac{\hbar}{4}\sqrt{\dfrac{k}{m}} + \dfrac{\hbar}{4}\sqrt{\dfrac{k}{m}}\\\\&= \dfrac{\hbar}{2}\sqrt{\dfrac{k}{m}}\end{align*}

    This is identical to the ground state energy derived from the HO Schrödinger's Equation.

    Question 84:

    Consider an infinite square well with wall boundaries x=0 and x=L. Explain why the function ψ(x)=Acoskx is not a solution to the stationary Schrӧdinger equation for the particle in a box.

    The potential energy of the particle in a box is given by:

    \begin{equation} V(x)=\begin{cases} 
    0 & 0\leq x\leq L \\ 
    \infty & x< 0 \; and\; x> L \end{cases}\end{equation}

    The particle cannot penetrate into an area with infinite potential energy, meaning that:

    \begin{equation*} \Psi(x) = 0\ for\ x < 0\ and\ x > L \end{equation*}

    Because of the continuity requirement of the wavefunction, it is required that:

    \begin{equation*} \Psi(x) = 0\ for\ x = 0\ and\ x = L \end{equation*}

    The proposed wavefunction does not meet this requirement at x = 0.

    \begin{align*} \Psi(x) &= Acos(kx)\\\Psi(0) &= Acos(k*0)\\&= A\end{align*}

    The wavefunction will only be 0 at x=0 if A = 0. If this is the case, the wavefunction will be 0 everywhere, representing a particle with no kinetic energy. For this reason A=cos(kx) is not a solution to the Schrödinger equation.



    Question 8:

    How much does the power radiated by a blackbody increase when its temperature (in K) is tripled?

    The power radiated by a blackbody varies as the fourth power of the absolute temperature. When its temperature is tripled, the radiated power increases by a factor of 81.

    Question 16:

    The metals sodium, iron, and molybdenum have work functions 2.5 eV, 3.9 eV, and 4.2 eV, respectively. Which of these metals will emit photoelectrons when illuminated with 400 nm light?

    Only sodium will emit photoelectrons when illuminated with this wavelength of light.

    Question 31:

    If, in a hydrogen atom, an electron moves to an orbit with a larger radius, does the energy of the hydrogen atom increase or decrease?

    The energy of the hydrogen atom increases.

    Extra Credit 17 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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