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2016 Midterm 2 Solutions

  • Page ID
    120338
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    Q1

    The angular momentum is given by

    \[L = \hbar\sqrt{l(l+1)} \]

    The total number of nodes is

    \[Nodes = n-1 \]

    The number of angular nodes

    \[Angular Nodes = l \]

    and the radial nodes are what's left

    \[Radial Nodes = n - 1 - l \]

    For each orbital we have

    a. 1s - \(L = 0\) and \(Nodes = 0\)

    b. 2s - \(L = 0\) and \(Nodes = 1\) and \(Radial Nodes = 1\)

    c. 2p - \(L = \hbar\sqrt{1(1+1)} = \hbar\sqrt{2} \) and \(Nodes = 1\) and \(Angular Nodes = 1\)

    d. 3d - \(L = \hbar\sqrt{2(2+11} = \hbar\sqrt{6} \) and \(Nodes = 2\) and \(Angular Nodes = 2\)

    e. 5f - \(L = \hbar\sqrt{3(3+11} = \hbar\sqrt{12} \) and \(Nodes = 4\) and \(Angular Nodes = 3\) and \(Radial Nodes = 1\)

    Q2

    The tricky part is this question is realizing that only some of the atoms in the reaction have electrons

    \(He^{2+}\) (No electrons, E = 0) \( + H\) (Electrons E ≠ 0) \(\rightarrow He^+ \) (Electrons E ≠ 0) \( + H^+ \) ( No Electrons E = 0)

    For Hydrogen like atoms:

    \[E_n = \dfrac{-RZ^2}{n^2} \]

    \[\Delta E = E_{products} - E_{reactants} = \dfrac{-R \times 2^2}{1} - \dfrac{-R \times 1^2}{1} = -4R - -R = - 3R \]

    Q3

    Ziliang do you want to put in the full solution with the derivatives? That what the students will most likely do.

    An easier way to do the problem using the properties of commutators is:

    \[\big[\hat{L}_{x},\hat{L}_{z}\big]=\big[\big(y\hat{p}_{z}-z\hat{p}_{y} \big),\big(x\hat{p}_{y}-y\hat{p}_{x} \big)\big]\]

    \[=\big[y\hat{p}_{z},x\hat{p}_{y}\big] + \big[z\hat{p}_{y},y\hat{p}_{x}\big] - \big[y\hat{p}_{z},y\hat{p}_{x}\big] - \big[z\hat{p}_{y},x\hat{p}_{y}\big]\]

    \[=x\big[y,\hat{p}_{y}\big] \hat{p}_{z}+z\big[\hat{p}_{y},y\big] \hat{p}_{x}-0-0\]

    \[=x(i\hbar)\hat{p}_{z}+z(-i\hbar)\hat{p}_{x}=-i\hbar\big(z\hat{p}_{x}-x\hat{p}_z \big)=-i\hbar\hat{L}_y\]

    Q4

    a.

    You can find the bond length in a diatomic molecule from the rotational constant:

    \[B = \dfrac{\hbar^2}{2I} = \dfrac{h^2}{8\pi^2 \mu r^2} \]

    Or in wavenumbers \(cm^{-1} \)

    \[\tilde{B} = \dfrac {h}{ 800 \pi^2 c \mu r^2}\]

    So the radius or bond length is:

    \[r = \sqrt{\dfrac{h}{800 \pi^2 c \mu B}} \]

    And the value is

    \[r = .16 nm \]

    b.

    The force constant comes from the fundamental vibration frequency.

    \[\tilde{\nu} = \dfrac{1}{200 c \pi} \sqrt{\dfrac{k}{mu}} \]

    \[k = \mu ( 200 \tilde{\nu} c \pi )^2 \]

    \[k = 101 N/m \]

    c.

    The energy for the transition in a rotational model from a higher state to a lower state.

    \[\Delta E_J = B(J+1)(J+2) - BJ(J+1) = 2B(J+1) \]

    \[\Delta E_J = 2B(2+1) = 6B = 45.078 cm^{-1} \]

    This energy is absorbed since you are making an upward transition in the question. J=2 to J=3

    Q5

    \[\langle r \rangle=\int^\infty_0 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}}\times r \times 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}} \times r^2dr \tag{1}\]

    The final \(r^2 dr \) is the Jacobian in spherical coordinates if you use radial \(R_{10}(r)\) wavefunction for calculation as shown in \((1)\). You can get the same result with \(\psi_{100}\) wavefunction while using \(4\pi r^2 dr\) as Jacobian within the integral.
    \[4 \left (\dfrac {Z}{a_0} \right )^3 \int^\infty_0 e^{\dfrac{-2Zr}{a_0}}\times r^3dr \tag{2}\]
    Using the great integral:
    \[ \int^\infty_0 x^ne^{-bx} dx = \dfrac{n!}{b^{n+1}} \tag{3} \]
    The answer appears
    \[4 \left (\dfrac {Z}{a_0} \right )^3 \times 3! \times \dfrac{a_0^4}{16Z^4} \tag{4}\]
    \[ \langle r \rangle = \dfrac{3a_0}{2Z} \tag{5}\]
    \[\langle r^2 \rangle=\int^\infty_0 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}}\times r^2 \times 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}} \times r^2dr \tag{6}\]
    \[4 \left (\dfrac {Z}{a_0} \right )^3 \int^\infty_0 e^{\dfrac{-2Zr}{a_0}}\times r^4dr \tag{7}\]
    Using the great integral \((3)\).:
    \[4 \left (\dfrac {Z}{a_0} \right )^3 \times 4! \times \dfrac{a_0^5}{32Z^5} \tag{8}\]
    \[ \langle r^2 \rangle = \dfrac{3a_0^2}{Z^2} \tag{9}\]

    Q6

    The Hamiltonian is the same as the pariticle in the box:

    \[\hat{H} = \dfrac{-\hbar^2}{2m} \dfrac{d^2}{dx^2} + 0 \]

    And the variational wavefunction is:

    \[\Phi = x(x-L) = x^2 - Lx \]

    So the variational energy is:

    \[E = \dfrac{\langle \Phi | \hat{H} | \Phi \rangle}{\langle \Phi | \Phi \rangle} \]

    Find \( \hat{H} |\Phi \rangle \) first.

    \[ \hat{H} |\Phi \rangle = \dfrac{-\hbar^2}{2m} \dfrac{d^2}{dx^2} (x^2 - xL) = \dfrac{-\hbar^2}{2m} \dfrac{d}{dx} (2x-L) = \dfrac{-\hbar^2}{2m} 2 \]

    Then the numerator

    \[\langle \Phi | \hat{H} | \Phi \rangle = \dfrac{-\hbar^2}{m} \langle 2x^2 - 2xL \rangle = \dfrac{-\hbar^2}{m} \int_0^L 2x^2 - 2xL dx \]

    \[\langle \Phi | \hat{H} | \Phi \rangle = \dfrac{-\hbar^2}{2m} 2/3 x^3 - x^2 L \Big|_0^L = \dfrac{\hbar^2L^3}{6m} \]

    The denominator

    \[\langle \Phi | \Phi \rangle = \langle (x^2 -xL)(x^2 - xL) \rangle = \int_0^L x^4 - 2x^3L + x^2L^2 = \dfrac{x^5}{5} - \dfrac{x^4L}{2} + \dfrac{x^3L^2}{3} \Big|_0^L \]

    \[\langle \Phi | \Phi \rangle = \dfrac{L^5}{30} \]

    Combine the two pieces

    \[E = \dfrac{\hbar^2L^3}{6m} \dfrac{30}{L^5} = \dfrac{5\hbar^2}{mL^2} = \dfrac{5h^2}{4\pi^2mL^2} \]

    the correct energy for the particle in a box is:

    \[E_{True} = \dfrac{h^2}{8mL^2} \]

    Which means the variation approximation is bigger by a factor of:

    \[\dfrac{10}{\pi^2} = 1.013\] or 1.3%


    2016 Midterm 2 Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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