# Quantum Chemistry/Printable version

Quantum Chemistry

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# Example 1

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle for a quantum harmonic oscillator in the ground state (n=0).

## Find ⟨x⟩, ⟨x2⟩, ⟨px⟩ and ⟨px2⟩ for a quantum harmonic oscillator in the ground state, then determine the uncertainty on the position and momentum. Is the product of the uncertainty on position and momentum consistent with the Heisenberg's Uncertainty Principle?

 Heisenberg's Uncertainty Principle $\qquad \delta _{x}\cdot \delta _{p}\geq {\frac {\hbar }{2}}\qquad {\text{where}}\ \hbar ={\frac {h}{2\pi }}$ The wavefunction of a quantum harmonic oscillator in the ground state is:

$\qquad \qquad \Psi _{0}(x)={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}$  Using this wavefunction the average position and the average of the square of the position can be calculated.

The average position:

$\qquad \qquad \qquad \qquad \langle x\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x\cdot \Psi _{0}(x)\ dx$

$\qquad \qquad \qquad \qquad \quad \ =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x\cdot e^{\frac {-2\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}\ dx$

 use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$ $\qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$

$\qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$

$\qquad \qquad \qquad \qquad \langle x\rangle =0$

The average square of the position:

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x^{2}\cdot \Psi _{0}(x)\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x^{2}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx$

 use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}$

 use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl [}1-(-1){\biggl ]}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl [}2{\bigl ]}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle ={\frac {1}{2\alpha }}$

The uncertainty on the position:

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ ={\sqrt {{\frac {1}{2\alpha }}-0}}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\frac {1}{2\alpha }}}$

The average momentum:

$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}\cdot \Psi _{0}(x)\ dx\qquad \qquad \qquad \qquad \qquad \qquad {\text{where}}\ {\widehat {p}}_{x}=-i\hbar {\Bigl (}{\frac {\partial }{\partial x}}{\Bigr )}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot -i\hbar {\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}-\alpha x\ e^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\cdot x\ e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }x\ e^{-\alpha x^{2}}\ dx$

 use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl [}{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl [}{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$

$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =0$

The average square of the momentum:

$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}^{2}\cdot \Psi _{0}(x)\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot {\Biggl (}-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}i^{2}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}(-1)\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ {\biggl (}-\alpha xe^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}e^{\frac {-\alpha x^{2}}{2}}+\alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}{\biggl )}\ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\ \alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$

 use $\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx={\sqrt {\frac {\pi }{\alpha }}}$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha \int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx{\biggl )}$

 use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl [}{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}$

 use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl (}{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}$

$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle ={\frac {1}{2}}\alpha \hbar ^{2}$

The uncertainty on the momentum:

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\langle p_{x}^{2}\rangle -\langle p_{x}\rangle ^{2}}}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ={\sqrt {{\frac {a\hbar ^{2}}{2}}-0^{2}}}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\frac {a^{2}}{2}}}\ \hbar$

The product of the uncertainty on the position and the uncertainty on the momentum is:

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta _{x}\cdot \delta p_{x}={\frac {1}{\sqrt {2\alpha }}}\cdot {\sqrt {\frac {\alpha }{2}}}\ \hbar$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar$

This is equal to ${\frac {\hbar }{2}}$ , therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.

# Example 2

The correspondence principle for the particle in a 1D box

## Question

Starting with the wave equation of a 1D box:

$\Psi \left(x\right)={\sqrt {\frac {2}{L}}}sin\left({\frac {n\pi }{L}}x\right)$

Show that the average probability of the particle in a 1D box follows the correspondence principle given that the average probability according to classical mechanics is:

$P\left(x\right)_{classical}={\frac {1}{L}}$

Where $L$  is the length of the 1D box, $n$  is the principle quantum number, and $x$  is the position of the particle in a 1D box.

## Solution

The probability distribution of a particle in a 1D box is represented by the wavefunction multiplied by it's complex conjugate $\Psi ^{*}$  over the full length of the box. The probability of finding a particle in a specific range is determined by integrating the wavefunction multiplied by it's complex conjugate over a distance between two given values:

$P\left(x\right)=\int _{a}^{b}\Psi ^{*}(x)\Psi (x)$

$\left\vert \Psi ^{2}\right\vert ={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)\cdot {\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)$

$\left\vert \Psi ^{2}\right\vert ={\frac {2}{L}}sin^{2}\left({\frac {n\pi }{L}}x\right)$

The correspondence principle states that as quantum numbers become large, quantum mechanics reproduces expected results from classical mechanics. Therefore, the average probability of finding a particle in a 1D box for quantum mechanics should match classical mechanics as the quantum number reaches infinity according to the correspondence principle.

The average value of an integrand is given by the formula:

$f_{avg}\left(x\right)={\frac {1}{b-a}}\int _{a}^{b}f\left(x\right)dx$

In this example, the function to be integrated is a $\sin ^{2}x$  function, which is a function with continuously repeating cycles from $0$  to $\pi$ . Therefore, determining the average over one cycle determines the average over an infinite amount of cycles, going to infinity represents the principle quantum $n$  going to very large numbers. So, the average of $\sin ^{2}x$  as $n$  goes to infinity is determined between one cycle from $0$  to $\pi$ .

${\overline {\sin ^{2}\left(x\right)}}={\frac {1}{\pi -0}}\int _{0}^{\pi }\sin ^{2}\left(x\right)dx$ , as ${x}\rightarrow \infty$

Using the trigonometric relationships below, the solution to the original integral becomes trivial.

$\cos ^{2}x=1-\sin ^{2}x$

$\cos {2x}=\cos ^{2}x-\sin ^{2}x$

$\sin ^{2}x={\frac {1}{2}}-{\frac {\cos {2x}}{2}}$

The new value of $\sin ^{2}x$  is inserted into the integral and solved for.

${\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\int _{0}^{\pi }\left[{\frac {1}{2}}-{\frac {\cos {2x}}{2}}\right]dx$  , as ${x}\rightarrow \infty$

${\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\left({\frac {x}{2}}-{\frac {\sin {2x}}{4}}\right){\Big |}_{0}^{\pi }$ , as ${x}\rightarrow \infty$

${\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{\pi }}\left({\frac {\pi }{2}}-0\right)$ , as ${x}\rightarrow \infty$

${\overline {\sin ^{2}{\left(x\right)}}}={\frac {1}{2}}$ , as ${x}\rightarrow \infty$

Thus, the average value of $\sin ^{2}x$  as the principle quantum number goes to infinity is equal to ${\frac {1}{2}}$ . By plugging that value into the probability distribution formula for a particle in a 1D box, the average probability becomes ${\frac {1}{L}}$ .

$\left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {2}{L}}\sin ^{2}\left({\frac {n\pi }{L}}x\right)$ , as ${n}\rightarrow \infty$

$\left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {2}{L}}\cdot {\frac {1}{2}}$

$\left\vert \Psi ^{2}\left(x\right)\right\vert ={\frac {1}{L}}$

This matches the average probability of a particle in a 1D box for classical mechanics as given in the question and demonstrates the correspondence principle using the 1D box model.

# Example 3

Write a question about calculating the number of nodes in a particle in a 1D box wavefunction.

## Question 3

Calculate the number of nodes for a particle in a 1D box when n=2 and n=3 when the length of the box is L=5, and give the x-intercept of the node(s).

 Wavefunction for a particle in a 1D-Box $\psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$ #### Part 1

nnodes=n-1

nnodes=2-1=1

We're looking for one x-intercept.

$0\div {\sqrt {\frac {2}{5}}}={\sqrt {\frac {2}{5}}}\sin \left({\frac {2\pi x}{5}}\right)\div {\sqrt {\frac {2}{5}}}$

$0={\sqrt {\frac {2}{5}}}\sin \left({\frac {2\pi x}{5}}\right)$

Use the form $\sin(kx)$  where $k=\left({\frac {2\pi }{5}}\right)$

$Period=\left({\frac {2\pi }{k}}\right)$

$Period=\left({\frac {2\pi }{\frac {2\pi }{5}}}\right)=5$

$\left({\frac {Period}{2}}\right)$  is the x-intercept

x-intercepts are 0, $\left({\frac {5}{2}}\right)$ , 5

0 and 5 are the edges of the box and $\left({\frac {5}{2}}\right)$  is the only node.

#### Part 2

nnodes=n-1

nnodes=3-1=2

We're looking for two x-intercepts.

$0\div {\sqrt {\frac {3}{5}}}={\sqrt {\frac {3}{5}}}\sin \left({\frac {3\pi x}{5}}\right)\div {\sqrt {\frac {3}{5}}}$

$0={\sqrt {\frac {3}{5}}}\sin \left({\frac {3\pi x}{5}}\right)$

$k=\left({\frac {3\pi }{5}}\right)$  $Period=\left({\frac {3\pi }{\frac {3\pi }{5}}}\right)={\frac {10}{3}}$

x-intercepts are $\left({\frac {5n}{3}}\right)$

x-intercepts = 0, $\left({\frac {5}{3}}\right),\left({\frac {10}{3}}\right),5$

0 and 5 are the edges of the box, $\left({\frac {5}{3}}\right)$  and $\left({\frac {10}{3}}\right)$  are the nodes.

# Example 4

## Example 4

Write a question about calculating the frequency of a photon to calculate the energy to transition between two levels of an electron in a 1D box

## Question

An electron in a 1D box emits a photon as the electron transitions to a lower energy level. If the length of the 1D box is equal to 1.0 cm, and the quantum number transition is $n_{{3}\rightarrow {2}}$ , what is the electromagnetic radiation frequency of the emitted photon?

Solution:

The energy level of a particle in a 1D box at a specific quantum number (${E_{n}}$ ) is,

$E_{n}={\frac {h^{2}}{8mL^{2}}}{n^{2}}$

Where $h$  is equal to Planck's constant (6.62607015 x 10-34 J$\cdot$ s), $n$  is equal to the quantum number ($n$  = 1, 2, 3, ...), $m$  is equal to the mass of the particle, and $L$  is equal to the length of the 1D box. For an electron, the mass is equal to 9.10938356 x 10-31 kg.

Since ${E_{n}}\propto {n^{2}}$  (assuming the mass and box length are constant), the energy level increases by a factor of 4 as the quantum number increases by a factor of 2. Therefore, if a particle in a 1D box undergoes an energy level transition, there is a difference between the initial and final quantum number energy levels. The energy level difference ($\Delta {E}$ ) of a particle in a 1D box that has undergone an energy level transition is,

$\Delta {E}={E_{n_{f}}}-{E_{n_{i}}}$

$=\left[{\frac {h^{2}}{8mL^{2}}}{n_{f}^{2}}\right]-\left[{\frac {h^{2}}{8mL^{2}}}{n_{i}^{2}}\right]$

$\Delta {E}={\frac {h^{2}}{8mL^{2}}}{({n_{f}^{2}}-{n_{i}^{2}}})$

Where $n_{f}$  is equal to the final quantum number, and $n_{i}$  is equal to the initial quantum number.

If $n_{f}>n_{i}$ , $\Delta {E}$  is a positive value; photon absorbed.

If $n_{f} , $\Delta {E}$  is a negative value; photon emitted.

Therefore, the energy level difference of an electron in a 1D box with a length of 1.0 cm, which has undergone a $n_{{3}\rightarrow {2}}$  transition is,

$\Delta {E}=\left[{\frac {(6.6261\times 10^{-34}{\text{ J}}{\text{ s}})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({2}^{2}-{3}^{2}})=\left[{\frac {(6.6261\times 10^{-34}{\text{ m}}^{2}{\text{ kg}}{\text{ s}}^{-1})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({2}^{2}-{3}^{2}})$

$\Delta {E}=-3.01\times 10^{-33}{\text{ J}}$

The energy level difference for the electron which underwent a $n_{{3}\rightarrow {2}}$  transition is equal to -3.01 × 10-33 J. Since $n_{f} , 3.01 × 10-33 J was emitted from the electron. If the electron underwent a $n_{{2}\rightarrow {3}}$  transition, the electron would absorb the same amount of energy that was emitted from the $n_{{3}\rightarrow {2}}$  transition which was 3.01 × 10-33 J.

$\Delta {E}=\left[{\frac {(6.6261\times 10^{-34}{\text{ m}}^{2}{\text{ kg}}{\text{ s}}^{-1})^{2}}{8(9.1094\times 10^{-31}{\text{ kg}})(1.0\times 10^{-2}{\text{ m}})^{2}}}\right]{({3}^{2}-{2}^{2}})=3.01\times 10^{-33}{\text{ J}}$

Therefore,

$E_{{\text{photon}}_{n_{2\rightarrow 3}}}=E_{{\text{photon}}_{n_{3\rightarrow 2}}}$

The energy of a photon has a specific frequency of electromagnetic (EM) radiation, and the energy is directly proportional to the frequency. The energy of the photon is equal to,

$E=h\cdot {f}$

Where $h$  is equal to Planck's constant (6.62607015 x 10-34 J$\cdot$ s), and $f$  is equal to EM radiation frequency.

Rearranging this equation allows for the calculation of the photon EM radiation frequency,

$E=h\cdot {f}$

$f={\frac {E}{h}}$

The calculated photon energy was equal to 3.01 × 10-33 J, therefore the EM radiation frequency of the emitted photon from the $n_{{3}\rightarrow {2}}$  transition of an electron in a 1D box with a length of 1.0 cm is,

$f={\frac {3.01\times 10^{-33}{\text{ J}}}{6.6261\times 10^{-34}{\text{ J}}{\text{ s}}}}$

$f=4.54{\text{s}}^{-1}=4.54{\text{Hz}}$

# Example 5

For a particle (assume the particle is an electron with 1 quantum number.) in a 1D box of length 5 cm, the equation of energy levels of a particle in a 1D box is,

$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$

a. If the length of the 1D box is increased to 10 cm, what is the change in the energy level of this particle in the box?

b. If the length of the 1D box is decreased to 2 cm, what is the change in the energy level of this particle in the box?

c. Explain the effect of length changes on the energy levels of particles in a 1D box.

## Solutions

The energy levels of a particle in 1D box:

$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$

where h is Planck's constant equal to $6.626\times 10^{-34}{\text{ J s}}$

Because this particle is an electron with 1 quantum number:

m is the mass of this particle equal to $9.109\times 10^{-31}{\text{ kg}}$

n is the quantum number of the particle equal to 1

L is the length of this 1D box equal to $5{\text{ cm}}$

for this question,

$\Delta {E}=E_{nf}-E_{ni}$

$=\left[{\frac {h^{2}n^{2}}{8m}}{\frac {1}{L_{f}^{2}}}\right]-\left[{\frac {h^{2}n^{2}}{8m}}{\frac {1}{L_{i}^{2}}}\right]$

$=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$

a. for part a, the initial length of this 1D box: $L_{i}=5{\text{ cm}}=5\times 10^{-2}{\text{ m}}$ , the final length of this 1D box: $L_{f}=10{\text{ cm}}=10\times 10^{-2}{\text{ m}}$

$\Delta {E}=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$

$=\left[{\frac {(6.626\times 10^{-34}{\text{ J s}})^{2}(1)^{2}}{(8)(9.109\times 10^{-31}{\text{ kg}})}}\right]\left[{\frac {1}{(10\times 10^{-2}{\text{m}})^{2}}}-{\frac {1}{(5\times 10^{-2}{\text{m}})^{2}}}\right]$

$=-1.807\times 10^{-35}{\text{ J}}$

b. for part b, the initial length of this 1D box: $L_{i}=5{\text{ cm}}=5\times 10^{-2}{\text{ m}}$ , the final length of this 1D box: $L_{f}=2{\text{ cm}}=2\times 10^{-2}{\text{ m}}$

$\Delta {E}=\left[{\frac {h^{2}n^{2}}{8m}}\right]\times \left[{\frac {1}{L_{f}^{2}}}-{\frac {1}{L_{i}^{2}}}\right]$

$=\left[{\frac {(6.626\times 10^{-34}{\text{ J s}})^{2}(1)^{2}}{(8)(9.109\times 10^{-31}{\text{ kg}})}}\right]\left[{\frac {1}{(2\times 10^{-2}{\text{m}})^{2}}}-{\frac {1}{(5\times 10^{-2}{\text{m}})^{2}}}\right]$

$=1.265\times 10^{-34}{\text{ J}}$

c. Based on the calculations in part a and b, the energy level of the particles in the 1D box is decreased by the increase of this 1D box length, and the energy level of the particles in the 1D box is increased by the decrease of this 1D box length. Therefore, the energy level of the particles in the 1D box is negatively related to the length of this 1D box.

# Example 6

Write an example question comparing the ground state energy of H, He+ and Li2+

# Example 7

## Example Problem 7

Using calculus, derive the most probable radius of an electron 1s orbital.

### Solution

To determine the equation for the most probable radius of an electron in an 1s orbital we must calculate the value of r at the maximum point on the probability distribution. This is done by setting the derivative of radial probability to zero and solving for radius.

The general formula for the radial probability is.

${\text{Radial Probabilty}}=P(r)=R_{nl}^{*}(r)R_{nl}(r)\cdot r^{2}=(R_{nl}(r))^{2}\cdot r^{2}$ 

After inserting the wavefunction of a 1s orbital we can simplify the radial probability equation as follows.

 Radial Component of Wave Function of 1s Orbital $R_{10}(r)=2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}$ $P_{1s}(r)=\left(2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}\right)^{2}\cdot r^{2}$

$P_{1s}(r)=4\left({\frac {Z}{a_{o}}}\right)^{3}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}$

$P_{1s}(r)={\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}$

The derivative of the simplified probability function must then be calculated.

${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {\delta }{\delta r}}\left({\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)$

The constants can be excluded from the rest of the equation by moving them to the front of the derivative.

${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}{\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)$

We see that the function is composed of the product of two smaller functions of the radius, multiplied by constants. Therefore we can apply the derivative product rule to solve for the equation of the derivative of the probability function.

 Product Rule of Derivatives ${\frac {\delta }{\delta r}}\left(f(x)\cdot g(x)\right)={\frac {\delta }{\delta r}}(f(x))\cdot g(x)+f(x)\cdot {\frac {\delta }{\delta r}}(g(x))$ ${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left({\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot {\frac {\delta }{\delta r}}r^{2}\right)\right)$

${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left(\left({\frac {-2Z}{a_{o}}}\right)\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot 2r\right)\right)$

Simplifying and moving the constants now will be beneficial to simplifying the final equation

${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)$

${\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)=0={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)$

After setting the derivative of radial probability to zero as seen above it can be seen that the only way that the equation can be equal to zero is if the polynomial part of the formula is equal to zero. This is the case because $e^{x}\neq 0$  and the first part of the equation is constant which means it cannot be zero. Therefore... $0=\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r$

The equation for the exact value of the most probable radius in a 1s orbital is then calculated using the quadradic equation.

 Quadratic Equation $x={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}$ Therefore let $x=r$ , $a={\frac {-Z}{a_{o}}}$ , $b=1$  and $c=0$  $r={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}$

$r={\frac {-(1)\pm \left((1)^{2}-4\left({\frac {-Z}{a_{o}}}\right)(0)\right)^{\frac {1}{2}}}{2\left({\frac {-Z}{a_{o}}}\right)}}$

$r={\frac {-1\pm \left(1-0\right)^{\frac {1}{2}}}{\left({\frac {-2Z}{a_{o}}}\right)}}$

$r={\frac {-1\pm 1}{\frac {-2Z}{a_{o}}}}$

The most probable radius cannot be a negative distance or zero meaning the numerator must be negative so that it is cancelled out by the negative value of the denominator. This means that the plus or minus operator in the numerator must be minus. $r={\frac {-1-1}{\frac {-2Z}{a_{o}}}}$

$r={\frac {1}{\frac {Z}{a_{o}}}}$

$r=r_{\text{most probable}}={\frac {a_{o}}{Z}}$

In conclusion the most probable radius for an electron in an 1s orbital is $r_{\text{most probable}}={\frac {a_{o}}{Z}}$  where that radius is proportional to the bohr radius $\left(a_{o}={\frac {4\pi \epsilon _{o}\hbar ^{2}}{m_{e}e^{2}}}=5.2917721090\cdot 10^{-11}m\right)$  and inversely proportional to the nuclear charge of the nucleus$\left(Z={\text{number of protons in nucleus of atom}}\right)$ . This means that for an atom with one proton like hydrogen [H] the most probable radius is $1\cdot a_{o}=5.2917721090\cdot 10^{-11}m$  while for the helium ion [He+] which also only has one electron in the 1s orbital the most probable radius is $0.5\cdot a_{o}=2.645886055\cdot 10^{-11}m$ .

# Example 8

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle holds for the J=0 state of a quantum rigid rotator

## Background Information

### Rigid rotor model

In quantum chemistry, the rigid rotor model is used to describe the rotations in molecules, such as HCl. The assumptions used in the rigid rotor model is that the rotating molecules is rigid, and the changes in bond length that naturally occur in a molecule (such as vibrational) is insignificant compared to the bond length re, and thus, negligible. 

The total energy of the system is the summation of the potential energy and the kinetic energy. The potential energy of a rigid rotor is 0 given the assumption that the rigid rotor bond length is constant. As such, the total energy of the system is equal to KE, which is equal to the angular momentum.  The energy level (EJ) of a linear rigid rotor model (such as HCl) is given by the equation:

$E_{J}={\frac {\hbar }{2I}}J(J+1)$  

In which I is the inertia, based on the reduced mass of the diatomic and bond length, and J is the quantum energy level.

The rigid rotor model is 3 dimensional, and for ease of calculations, rather than 2 sets of masses that are used (m1 and m2), 1 reduced mass is used instead (μ). 

And in a spherical model of the rigid rotor, there are 2 variables that are used to determine position, the angels θ and ϕ, given that r is the constant bond length re.

Thus, the position of the reduced mass is given by the wave function:

$Y(\theta ,\varphi )$

### Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle states that the exact position and momentum of a particle cannot be determines at one given point, and the more precisely either is determined, the less certain the other would be.

However, even thou the exact position and momentum cannot be calculated at a given moment, they can be related, and that relation is: 

$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}$

In which $\hbar$  is the reduced Planck constant, $\bigtriangleup \phi$   is the uncertainty of position, and $\bigtriangleup L_{z}$  is the uncertainty of momentum.

However, in the rigid rotor model, this equation does not work. The Heisenberg inequality is recalculated: 

$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {1}{2}}|<[L_{z},\phi ]>|={\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$

$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$

## Example

#### Question: using the rigid rotor model at J=0, what is the moment of angular momentum and position? Does the rigid rotor follow Heisenberg’s principle at J=0?

In the ground state zero-point, J = 0, and thus, the energy level of the rigid rotor is 0.

Given that the energy is 0, the angular momentum is known (Lz = 0).

The probability density of the position can be obtained using the wavefunction:

Average of position = 0, as such, the variance is calculated:

$\delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}$

$<\varphi >=\int _{0}^{2\pi }(Y^{*})(\varphi )(Y)d\varphi$

Y = $i^{m+|m|}{\sqrt {{\frac {2J+1}{2}}{\frac {(J-|m|)!}{(J+|m|)!}}}}.P_{0}^{0}cos(\theta )$

at m=0 J=0

Y = ${\frac {1}{\sqrt {4\pi }}}$

$<\varphi >=\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi )({\frac {1}{\sqrt {4\pi }}})d\varphi$

$<\varphi >=({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi )d\varphi$

$<\varphi >=({\frac {1}{4\pi }})2\pi ^{2}={\frac {\pi }{2}}$

$<\varphi >^{2}={\frac {\pi ^{2}}{4}}$

$<\varphi ^{2}>=\int _{0}^{2\pi }(Y^{*})(\varphi )^{2}(Y)d\varphi$

$<\varphi ^{2}>=\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi ^{2})({\frac {1}{\sqrt {4\pi }}})d\varphi$

$<\varphi ^{2}>=({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi ^{2})d\varphi$

$<\varphi ^{2}>=({\frac {1}{4\pi }}){\frac {8\pi ^{3}}{3}}={\frac {2\pi ^{2}}{3}}$

$\delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}={\sqrt {{\frac {2\pi ^{2}}{3}}-{\frac {\pi ^{2}}{4}}}}=\pm {\frac {\sqrt {15}}{6}}\pi$

if $\delta \phi$  is plotted on a sphere

Where m = integers between J and – J.

|m| = number of longitudinal node

J = number of latitudinal nodes

At ground state, J = 0, |m| = 0. Thus, probability of position is equally spread across the sphere.

at J=0,

${\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]=0$

thus

$\bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]$

$(\pm {\frac {\sqrt {15}}{6}}\pi )(0)\geq 0$

is satisfied

Thus, this shows that at ground state, J=0, the rigid rotor model follows the Heisenberg uncertainty.

# Example 9

Write a question and its solution that calculates the locations of the nodes of an electron in a 2s orbital.

## The Question

The question is to find the location of the radial node in a 2s electron for a hydrogen atom. To find the node one can start by analyzing, generally, how many nodes one should expect to see in a 2s electron system. There are two equations that give the number of nodes present in an orbital, the angular node equation and radial node equations:

$n-\ell -1=n_{r}$

2. Angular Nodes:

$\ell =n_{a}$

Therefore ℓ must be determined, and based on the table 1 data one can determine ℓ is equal to 0. And the n is equal to the $n$  which is 2, this comes from the number before the orbital type which tells you the principle quantum number.

Table 1: Orbitals and Quantum Number
Orbital Angular

Momentum

Quantum

Number (ℓ)

s 0
p 1
d 2
f 3

So how many nodes are there?

First analyze the number of angular nodes:

$\ell =n_{a}$

$0=n_{a}$

Therefore, the number of angular nodes is 0.

$n_{r}=n-\ell -1$

$n_{r}=2-0-1$

$n_{r}=1$

Therefore there is one radial node present in a 2s orbital, resulting in the question becoming where is the location of that radial node?

The Wavefunction

Now the next main step is to determine what wavefunction describes the wavefunction this scenario of the 2s electron. The wavefunction can be found online which is$:^{1}$

$\psi _{2s}={\frac {1}{4(2\pi )^{\frac {1}{2}}}}\left({\frac {1}{a_{0}}}\right)^{3/2}\left(2-{\frac {r}{a_{0}}}\right){\text{e}}^{-r/(2a_{0})}$

In the equation the $\psi _{2s}$  is the wavefunction for the 2s electron, 𝒓 is the radius, and $a_{0}$  is the Bohr radius.

Based on the equation we can then solve for the position of the electron. The best way to do this is to find where is the equation going to be equal to zero and what term that contains the position causes this. The first part ${\frac {1}{4(2\pi )^{\frac {1}{2}}}}$  is a constant thus won't change with the radius, the position, of the electron. So the only place that will change with the position of the electron are the 𝒓 terms. With ${\text{e}}^{-r/(2a_{0})}$  the 𝒓 term can be any number and the term won't be zero unless the 𝒓 is approaching infinity, while the $\left(2-{\frac {r}{a_{0}}}\right)$  can potentially be equal to zero since it has 2 subtracted by the position term. Therefore one can set this term equal to zero and solve for 𝒓.

$0=2-{\frac {r}{a_{0}}}$

$2={\frac {r}{a_{0}}}$

$2a_{0}=r$

Therefore, we get the solution to the position of the radial node which is 𝒓 $=2a_{0}$ , so when 𝒓 $=2a_{0}$  the probability of the electron being there is 0 all around the nucleus creating a node. The $a_{0}$  has a length of 52.9 pm which means the node is 105.8 pm in radius away from the nucleus.

In Conclusion

In conclusion the approach to the problem of finding nodes for an electron in an orbital boils down to first finding the number of theoretical nodes, then determining the wavefunction, analyzing the wavefunction's variables and solving for 0. After this is all done you will have the solution to the radial node locations. Further problems can be solved as well, because they are follow up questions that are made easier to solve after finding the nodes, like the position of the electron in it's most probable state. This problems follows the solution of arranging the P=$\psi _{2s}$ $\left(\psi _{2s}^{*}\right)$ $\left(r^{2}\right)$  then finding the derivative of the wavefunction and then simplifying it. The final step is to find the zero points, where the 𝒓 is equal to zero which will give the most probable locations of the electron. The practical applications of finding the nodal locations can help with understanding how orbitals work which can help with making molecular orbital diagrams and SALCs that can be used to determine the way atoms and molecules bond. Other applications include understanding the energy levels of the bonds and orbitals to predict possible interactions between molecules and atoms, for research purposes and chemical engineering.

Reference

1) Branson, J. The Radial Wavefunction Solutions. https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html (accessed November 16, 2021).

By Dmitry Ivanov

# Example 10

## Question

Use Planck's radiation law to find the surface temperature of the Sun when its maximum intensity of EM radiation is emitted at 504 nm. Ensure the temperature units are in Kelvins.

### Getting the correct function

A blackbody material is defined by its ability to absorb ALL radiation that falls onto it. When the blackbody is at constant temperature, the distribution of its emission frequency can be determined by assuming its only direct relation is to temperature. Furthermore, the frequency of the electromagnetic (EM) radiation can also be measured in units of wavenumber ${\text{cm}}^{-1}$ .

{\begin{aligned}\rho (\nu ,T)&={\frac {2h\nu ^{2}}{c^{2}}}\left\langle E(\nu ,T)\right\rangle &&\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\left\langle E(\nu ,T)\right\rangle \end{aligned}}

Planck's approach to defining ${\textstyle \langle E(v,T)\rangle }$  was deriving for a closed form Harmonic Oscillator was based on Boltzmann's distribution. The resulting form of Planck's law can then be applied to the question now that all parameters are known except for the parameter in question. The model below shows the radiation intensity distribution (i.e., area of the curve) of EM with respect to frequency (in Hertz; Hz) and temperature (in Kelvins; K).

{\begin{aligned}d\rho (v,T)&=\rho (\nu ,T)d\nu {\text{, where }}\rho (\nu ,T)=\left({\frac {2h\nu ^{3}}{c^{2}}}\right)\left[\exp \left({\frac {h\nu }{k_{B}T}}\right)\right]^{-1}\\&{\text{Note that }}\left|{\frac {d\nu }{d\lambda }}\right|={\frac {c}{\lambda ^{2}}}\\d\rho (\lambda ,T)&=\rho (\lambda ,T)d\lambda {\text{, where }}\rho (\lambda ,T)=\left({\frac {2hc^{2}}{\lambda ^{5}}}\right)\left[\exp \left({\frac {hc}{k_{B}T\lambda }}\right)\right]^{-1}\end{aligned}}

Planck's law for a blackbody material predicts the behaviour of its quantitative properties (such as frequency, wavelength, and temperature) when the environment are in extreme conditions.
 Planck's Law of BlackBody Radiation $\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\times {\frac {1}{\left[\exp({\frac {hc}{k_{B}\lambda T}})\right]-1}}$ ### Solving for the temperature

The maximum value of the independent variable of any function can be found by equating the function's derivative to zero. Only $T$  is unknown and is with respect to the independent variable, $\lambda$ . Thus, Planck's law will be derived with respect to $\lambda$ .

First, for simplicity, let ${\textstyle x={\frac {hc}{k_{B}\lambda T}}}$ .

${d\rho \over d\lambda }=(2hc^{2})\left[{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\right]=0$

The first term consists only of non-zero constants so that leaves:

${\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}=0$

{\begin{aligned}{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}&={\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\\{\frac {xe^{x}}{e^{x}-1}}&=5\\{\frac {xe^{x}}{e^{x}-1}}-5&=0\\(x-5)e^{x}+5&=0\end{aligned}}

Subbing ${\textstyle x={\frac {hc}{k_{B}\lambda T}}}$  back in gives:

{\begin{aligned}\left({\frac {hc}{k_{B}\lambda T}}-5\right)e^{x}&=0&&\rightarrow &&&{\frac {hc}{k_{B}\lambda T}}-5=0\end{aligned}}

At max intensity, ${\textstyle \lambda _{max}=504{\text{nm}}}$ . Subbing in all known values and constants (in SI units) then rearranging to solve for T will determine the surface temperature of the Sun under these specific conditions.

Known Values
Variable Value Units
$h$  $6.62607\times 10^{-34}$  $m^{2}kg \over s$
$\lambda _{max}$  $504\times 10^{-9}$  $m$
$k_{B}$  $1.38065\times 10^{-23}$  ${\frac {m^{2}kg}{s^{2}K}}$
$c$  $2.99792\times 10^{8}$  $m \over s$

{\begin{aligned}T={\frac {hc}{5k_{B}\lambda _{max}}}={\frac {(6.62607\times 10^{-34})(2.99792\times 10^{8})}{5(1.38065\times 10^{-23})(504\times 10^{-9})}}=5709.432{\text{ }}K\end{aligned}}

Therefore, the surface temperature of the Sun at a maximum wavelength of 504 nm is just above 5709 Kelvins.

## References

1. a b Quantum Hypothesis Used for Blackbody Radiation Law https://chem.libretexts.org/@go/page/210776 (accessed Nov 20, 2021).
2. "Rotations and Vibrations of Polyatomic Molecules", Molecular Physics (Weinheim, Germany: Wiley-VCH Verlag GmbH): pp. 203–236, retrieved 2021-11-20
3. a b c

# Example 11

## Example 11

Use the 1D particle in a box model to estimate the wavelength of light required to excite an electron from a pi to pi* MO in ethene.

Solution:

The energy levels of a particle in a 1D box with a specific quantum number $n$ , are as follows.

$E_{n}={\frac {h^{2}n^{2}}{8mL^{2}}}$

In this equation $h$  represents Planck's constant, $m$  is the mass of the particle, and $L$  is the length of the box.

The pi electron in the double bond between the carbon atoms in ethene can be approximated to the particle in a 1D box model. This means that the mass of the particle in this question will be the mass of an electron, and the length of the box corresponds to the bond length between the carbon atoms in the molecule ethene.

Additionally, the energy equation above needs to be transformed into a equation for $\Delta E$  since the electron is moving from one energy level to another.

$\Delta E=E_{f}-E_{i}={\frac {h^{2}n_{i}^{2}}{8mL^{2}}}-{\frac {h^{2}n_{f}^{2}}{8mL^{2}}}={\frac {h^{2}}{8mL^{2}}}(n_{f}^{2}-n_{i}^{2})$

The change in energy between the pi and pi* MO in ethene can now be calculated knowing that the bond length between doubly bonded carbon atoms is 133pm and the mass of an electron is 9.1093856x10-31 kg. Moving from the ground state n=1 to an excited state of n=2 :

$\Delta E={\frac {h^{2}}{8mL^{2}}}(n_{f}^{2}-n_{i}^{2})$

$\Delta E={\frac {(6.626{\text{x}}10^{-34}{\frac {{\text{m}}^{2}{\text{kg}}}{\text{s}}})^{2}}{8(9.109\times 10^{-31}{\text{kg}})(133\times 10^{-12}{\text{m}})^{2}}}(2^{2}-1^{2})$

$\Delta E=1.0217\times 10^{-17}{\text{J}}$

Now that the energy required to excite the electron to the pi* orbital is known, the wavelength of light can be calculated through the following equation, where c is the speed of light in a vacuum and $\lambda$  is the wavelength of light.

$\Delta E={\frac {hc}{\lambda }}$

The equation can then be re-arranged to solve for the wavelength of light.

$\lambda ={\frac {hc}{\Delta E}}$

By plugging in the known constants and the value for $\Delta E$  that has been calculated above, the wavelength can be found.

$\lambda ={\frac {hc}{\Delta E}}$

$\lambda ={\frac {(6.626\times 10^{-34}{\frac {{\text{m}}^{2}{\text{kg}}}{\text{s}}})(2.998\times 10^{8}{\frac {\text{m}}{\text{s}}})}{1.0217\times 10^{-17}{\text{J}}}}$

$\lambda =1.9\times 10^{-7}{\text{m}}$

$\lambda =19{\text{nm}}$

Therefore the wavelength of light required to excite an electron from a pi to pi* molecular orbital in ethene is 19nm.

# Example 12

Write a question and its solution that shows the specific selection rule for a quantum rigid rotor.

Consider an N2 molecule with a bond length of 1.09 Å.

(a) Calculate the energy at the quantum number 3 using the specific selection rule for a rigid rotor.

Solution: This bond length is given (1.09 Å), and the reduced mass (μ) and the inertia (I) must be calculated to determine the energy at the angular momentum quantum number $J=3$ . The bond length and reduced mass must also be changed to SI units.

$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$ 

$\mu ={\frac {14.0067u\cdot 14.0067u}{14.0067u+14.0067u}}\cdot ({\frac {1.6605\times 10^{-27}kg}{1u}})$

$\mu =1.1629\times 10^{-26}kg$

Now, calculating the moment of inertia ($I$ ):

$I=\mu \cdot (r_{e})^{2}$ 

$=(1.1629\times 10^{-26}kg)\cdot (1.09\mathrm {\AA} \cdot 1.00\times 10^{-10}m)^{2}$

$=1.3816\times 10^{-46}kg\cdot m^{2}$

Since the molecule is linear (N2), the energy level ($E_{J}$ ) can be calculated with the linear rigid rotor equation:

$E_{J}=J(J+1)\cdot ({\frac {\hbar ^{2}}{2I}})$ 

$E_{J}=3(3+1)\cdot ({\frac {(1.05457\times 10^{-34}J\cdot s)^{2}}{2(1.3816\times 10^{-46}kg\cdot m^{2})}})$

$E_{J}=12\cdot (4.0247\times 10^{-23}J)$

$E_{J}=4.8296\times 10^{-22}J$

(b) Calculate the quantum number if the transition energy is 2.4176 × 10-22J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

The linear rigid rotor equation must be rearranged into linear form to solve for $J$ .

$E_{J}=J(J+1)\cdot ({\frac {\hbar ^{2}}{2I}})$

$J^{2}+J={\frac {2\cdot E\cdot I}{\hbar ^{2}}}$

$J^{2}+J-{\frac {2\cdot E\cdot I}{\hbar ^{2}}}=0$

$J^{2}+J-{\frac {(2)\cdot (2.4176\times 10^{-22}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}=0$

$J^{2}+J-6.0069=0$

To solve this relationship, the quadratic formula must be utilized:

$J={\frac {-b\pm {\sqrt {b^{2}-4\cdot a\cdot c}}}{2\cdot a}}$

$J={\frac {-1\pm {\sqrt {1^{2}-4\cdot 1\cdot -6.0069}}}{2\cdot 1}}$

$J={\frac {-1\pm {\sqrt {25.0275}}}{2}}$

$J={\frac {-1\pm {4.9973}}{2}}$

$J=2.0014,-3.0014$

Rigid rotor quantum numbers cannot be negative, ∴ $J=2$ .

This transition adheres to the specific selection rule for a quantum rigid rotor because the change in rotational quantum number is $\Delta J=\pm 1$ .

(c) Calculate the quantum number if the transition energy is 1.2073 × 10-21J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

The linear rigid rotor equation must be rearranged into linear form to solve for $J$ .

$J^{2}+J-{\frac {2\cdot E\cdot I}{\hbar ^{2}}}=0$

$J^{2}+J-{\frac {(2)\cdot (1.2073\times 10^{-21}J)\cdot (1.3816\times 10^{-46}kg\cdot m^{2})}{(1.05457\times 10^{-34}J\cdot s)^{2}}}=0$

$J^{2}+J-29.9970=0$

To solve this relationship, the quadratic formula must be utilized:

$J={\frac {-b\pm {\sqrt {b^{2}-4\cdot a\cdot c}}}{2\cdot a}}$

$J={\frac {-1\pm {\sqrt {1^{2}-4\cdot 1\cdot -29.9970}}}{2\cdot 1}}$

$J={\frac {-1\pm {\sqrt {120.98813}}}{2}}$

$J={\frac {-1\pm {10.9994}}{2}}$

$J=4.9997,-5.9997$

Rigid rotor quantum numbers cannot be negative, ∴ $J=5$ .This transition does not adhere to the specific selection rule for a quantum rigid rotor, because the change in rotational quantum number is not within the $\Delta J=\pm 1$ . Compared to part (a), it has a change of $\Delta J=\pm 2$ .

# Example 13

Write a question and its solution that shows the specific selection rule for a quantum harmonic oscillator

Calculate the energy for the vibrational transitions $v\rightarrow v+1$  and $v+6\rightarrow v+7$ . If they have the same energy gaps comment on why?

$\bigtriangleup Ev\rightarrow v+1$ $=hv_{0}{\Bigl (}(v+1)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl (}v+{\tfrac {1}{2}}{\Bigr )}$

$=hv_{0}{\biggl (}(v+1)+{\tfrac {1}{2}}-{\Bigl (}v+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$

$=hv_{0}{\biggl (}v+1+{\tfrac {1}{2}}-v-{\tfrac {1}{2}}{\biggr )}$

$=hv_{0}$

$\bigtriangleup Ev+1\rightarrow v+2$ $=hv_{0}{\Bigl (}(v+7)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl (}(v+6)+{\tfrac {1}{2}}{\Bigr )}$

$=hv_{0}{\biggl (}(v+7)+{\tfrac {1}{2}}-{\Bigl (}(v+6)+{\tfrac {1}{2}}{\Bigr )}{\biggr )}$

$=hv_{0}{\biggl (}v+7+{\tfrac {1}{2}}-v-6-{\tfrac {1}{2}}{\biggr )}$

$=hv_{0}$

The energy of the vibrational transitions from from $v\rightarrow v+1$  and $v+6\rightarrow v+7$  have the same energy gap $(hv_{0})$  . This is because of the specific selection rule for the quantum harmonic oscillator. The rule states that the molecule is only allowed to move up or down one vibrational energy level for the transition to occur. If the molecule diverges from the rule $\bigtriangleup V\pm 1$  then it is considered an overtone, and these are unlikely to occur.

# Example 14

Show using calculus the most probable position of a quantum harmonic oscillator in the ground state (n=0)

Question:

What is the most probable position of a quantum harmonic oscillator at the ground state? Calculate this using the probability density equation to find the most probable position at n=0.

 Probability distribution $P_{n}(x)=N_{n}^{2}H_{n}({\sqrt {\alpha }}x)^{2}e^{-\alpha x^{2}}$ Solution:

The Hermite polynomial at n=0 is:

$H_{0}(x)=1$

The normalization factor at n=0 is:

$N_{0}={\frac {1}{\sqrt {2^{1}1!}}}\left({\frac {\alpha }{\pi }}\right)^{1/4}=\left({\frac {\alpha }{\pi }}\right)^{1/4}$

α is a constant and is equal to:

$\alpha ={\sqrt {\frac {k\mu }{\hbar }}}$

The probability distribution at n=0:

$P_{n=0}(x)=\left({\frac {\alpha }{\pi }}\right)^{1/2}e^{-\alpha x^{2}}$

The most probable position is when the maximum probability distribution is:

${\frac {\partial P}{\partial x}}=0$

Applying this partial derivative to the probability distribution gives:

${\frac {\partial }{\partial x}}\left({\frac {\alpha }{\pi }}\right)^{1/2}e^{-\alpha x^{2}}=0$

The constants can be taken out of the derivative:

$\left({\frac {\alpha }{\pi }}\right)^{1/2}{\frac {\partial }{\partial x}}e^{-\alpha x^{2}}=0$

The derivative gives:

$\left({\frac {\alpha }{\pi }}\right)^{1/2}[-2\alpha xe^{-\alpha x^{2}}]=0$

Since it is equal to zero the constants can be divided out leaving:

$[-2\alpha xe^{-\alpha x^{2}}]=0$

Since all of the parts are multiplied they can be divided out leaving:

$x=0$

The point where the probability distribution is at a maximum for the ground state of n=0 for the quantum harmonic oscillator is 0.

# Example 15

Write an example question showing the determination of the bond length of CO using microwave spectroscopy

## Deriving the Required Equations

When a photon is absorbed by a polar diatomic molecule, such as carbon monoxide, the molecule can be excited rotationally. The energy levels of these excited states are quantized to be evenly spaced. The distance between each rotational absorption lines is defined as twice the rotational constant $({\tilde {\beta }})$  which can be measured via the following equation:

${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$ 

h = plank's constant = 6.626 ᛫10−34 J ᛫ s

c = speed of light = 2.998᛫108 m ᛫ s-1

I = moment of inertia

The energy to required to rotate a molecule around its axis is the moment of inertia $({\text{I}})$ . It can be calculated as the sum of the products the masses of the component atoms and their distance from the axis of rotation squared:

${\text{I}}=\sum _{i}m_{i}\cdot r_{i}^{2}$ 

Working it out for an heterogeneous diatomic molecule:

${\text{I}}=\left(m_{a}\cdot r_{a}^{2}\right)+\left(m_{b}\cdot r_{b}^{2}\right)$

${\text{I}}=r_{a}\cdot r_{b}\left(m_{a}+m_{b}\right)$

The distance from the atom to the center of mass $\left(r_{a}{\text{and}}r_{b}\right)$  cannot easily be measure, however, by setting the origin at the center of mass equation can be derive for the two values that uses the bond length $\left(r_{e}\right)$  as a variable:

$\left(m_{a}\cdot r_{a}\right)=\left(m_{b}\cdot r_{b}\right)=\left(m_{a}\cdot \left(r_{a}-r_{e}\right)\right)=\left(m_{b}\cdot \left(r_{b}-r_{e}\right)\right)$

$r_{a}={\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}$

$r_{b}={\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}$

Substituting these equations into the moment of inertia equation:

${\text{I}}={\biggl (}{\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot {\biggl (}{\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot \left(m_{a}+m_{b}\right)$

${\text{I}}={\biggl (}{\frac {m_{a}\cdot m_{b}\cdot r_{e}^{2}}{m_{a}+m_{b}}}{\biggr )}$

This equation can be simplified further if we imagine the rigid rotor as a single particle rotating around a fixed point a bond length away. The mass of this particle is the reduced mass $\left(\mu \right)$  of the two atoms that make up the diatomic molecule:

$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$ 

Simplifying the previous moment of inertia equation, we get:

${\text{I}}=\mu r_{e}^{2}$

## Solving for Bond Length

From here we have everything we need to be able to determine the bond length of a polar diatomic molecule such as carbon monoxide.

First, we must solve for the moment of inertia using the rotation constant:

${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$

${\text{I}}=\left({\frac {h}{8\pi ^{2}c{\tilde {\beta }}}}\right)$

As explained earlier, the rotational constant can be determined by measuring the distance between the rotational absorption lines and halfling it. In the case of ${\ce {^{12}C^{16}O}}$  the rotational constant is $193.1281$  m-1 . Plugging this value in we can determine the moment of inertia:

${\text{I}}=\left({\frac {6.626\cdot 10^{-34}}{8\pi ^{2}\cdot \left(2.998\cdot 10^{8}\right)\cdot \left(193.1281}\right)}}\right)$

${\text{I}}=1.44939\cdot 10^{-46}$  kg ᛫ m2

Now that we know inertia, we can rearrange the equation we derived earlier in order to determine the bond length:

${\text{I}}=\mu r_{e}^{2}$

$r_{e}={\sqrt {\frac {\text{I}}{\mu }}}$

$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{\mu }}}$

The exact atomic mass of ${\ce {^{12}C}}$  is 12.011 amu and ${\ce {^{16}O}}$  is 15.9994 amu . As such the reduce mass is calculated to be:

$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$

$\mu =\left({\frac {12.011\cdot 15.9994}{12.011+15.9994}}\right)$

$\mu =6.86062$  amu

$\mu =6.86062$  amu ᛫ $1.67377\times 10^{-27}\left({\frac {kg}{amu}}\right)$

$\mu =1.140\times 10^{-26}kg$

Plugging in the reduce mass back into our equation we can finally solve for the bond length of a carbon monoxide molecule:

$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{1.140\times 10^{-26}}}}$

$r_{e}={\sqrt {1.271\cdot 10^{-20}}}$

$r_{e}=1.12739\cdot 10^{-10}m$

$r_{e}=1.12739\mathrm {\AA}$

# Example 16

Write an example question showing the calculation of the frequency of EM radiation emitted when a HCl molecule transitions from the J=1–>0 rotational state.

Example 16: Giving the bond length re=1.27Å, find the frequency of the EM radiation emitted when a HCl molecule transitions from the J=1→0 rotational state

Given the reduced mass (μ) and inertia (I), the energy can be found and subsequently be used to find frequency of the transition.

Note that for rotational states for J=1→0, ΔE is not required as the transition states are in the ground state.

Inertia can be found by using reduced mass of the molecule and their bond length(re = 1.27Å):

$I=\mu \cdot r_{e}^{2}$

where reduced mass can be calculated by the following:

$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$ , where $m_{1}=1.00784u$  and $m_{2}=35.453u$

$\mu ={\frac {(1.00784)(35.453)}{1.00784+35.453}}=0.9770u$

Convert to SI units:

$1u=1.66054\times 10^{-27}kg$ , thus

$\mu =0.9799u\times {\frac {1.66054\times 10^{-27}kg}{1u}}=1.627\times 10^{-27}kg$

Note that the bond length must also be in SI units,

1Å= 1×10-10m

Bond length = 1.27×10-10m

Inertia can now be calculated:

$I=(1.627\times 10^{-27}kg)(1.27\times 10^{-10}m)^{2}=2.624\times 10^{-47}kg\cdot m^{2}$

After calculating the inertia, energy can be found:

$E={\frac {\hbar ^{2}}{2I}}J(J+1)=hv$

and $\hbar ={\frac {h}{2\pi }}$ , where h is the Planck's constant.

$E={\Biggl (}{\frac {6.626\times 10^{-34}m^{2}kgs^{-1}}{2\pi }}{\Biggr )}^{2}\times {\frac {1}{2\times 2.624\times 10^{-47}kg\cdot m^{2}}}\times 1(1+1)$

$E=4.240\times 10^{-22}J$

Frequency of the transmission can therefore be found:

$v={\frac {E}{h}}$

$v={\frac {4.240\times 10^{-22}J}{6.626\times 10^{-34}m^{2}kgs^{-1}}}=6.40\times 10^{11}s^{-1}$

# Example 17

The Bond constant of HCl is determined computationally to be 480 N/m. Given this information find the frequency of EM radiation required to excite the HCl molecule from its ground state to its first excited state.

## Solution

Given the bond constant of HCl(K), we use the relationship between fundamental frequency and bond constant to find the bond constant.

$\nu _{0}={\frac {1}{2\pi }}\left({\frac {K}{\mu }}\right)^{1/2}$

where $K$  is the bond constant.

$\mu$  is the reduced mass of HCl.

To find the reduced mass of HCl the masses of H and Cl are multiplied and divided by the sum of the masses.

$\mu ={\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}$

For HCl the reduced mass is calculated as

$\mu ={\frac {1.007842u\cdot 35.453u}{1.007842u+35.453u}}=0.979982u$

convert to the SI unit of Kg

$1u=1.66054\cdot 10^{-27}Kg$

$0.979983u=0.979983u\cdot 1.66054\cdot 10^{-27}Kg$

$\mu =1.6273\cdot 10^{-27}Kg$

To find the fundermental frequency

$\nu _{0}={\frac {1}{2\pi }}\left({\frac {K}{\mu }}\right)^{1/2}$

$={\frac {1}{2\pi }}\left({\frac {480N/m}{1.6273\cdot 10^{-27}Kg}}\right)^{1/2}$
$=8.6438\cdot 10^{13}hz$

After finding the fundamental frequency, the Energy at different quantum levels can be found by

$E_{v}=h\nu _{0}\left(v+{\frac {1}{2}}\right)$

For the ground state i.e. $v=0$

$E_{0}=h\nu _{0}\left(0+{\frac {1}{2}}\right)$
$E_{0}=h\nu _{0}\left({\frac {1}{2}}\right)$

For the first excited state i.e. $v=1$

$E_{1}=h\nu _{0}\left(1+{\frac {1}{2}}\right)$
$E_{1}=h\nu _{0}\left({\frac {3}{2}}\right)$

The difference in energy between the two states is

$\Delta E=E_{1}-E_{0}$
$\Delta E=h\nu _{0}\left({\frac {3}{2}}\right)-h\nu _{0}\left({\frac {1}{2}}\right)$
$\Delta E=h\nu _{0}$

and Energy is defined as Planck's constant multiplied by frequency

$\Delta E=hv$
$hv=h\nu _{0}$
$v=\nu _{0}=8.6438\cdot 10^{13}hz$

# Example 18

Write an example question showing the determination of the force constant of CO using IR spectroscopy.

## Question

Using IR spectroscopy, calculate the force constant of carbon monoxide (CO) and report the value in kN/m.

## Solution

To solve this question, the IR spectrum of CO is required. The website, webbook.nist.gov/chemistry/ is a database that is available to the public for access to IR spectra. If you are unfamiliar on acquiring an IR spectrum, follow these steps: with the nist website open, go to Search Option -> Name -> Enter: Carbon Monoxide -> Select: IR Spectrum -> Change Transmittance to Absorbance -> Finally, take a screen shot.

The CO IR spectrum earlier acquired is a low resolution rotational-vibrational spectrum. In IR spectrum's, the fundamental frequency can be determined by utilizing the P-branch, Q-branch, and R-branch. Remember that the Q-branch is pure vibrational, which is forbidden, therefore the peak does not exist and is located between the P-branch and R-branch.

Using the website, apps.automeris.io/wpd/ will allow for an accurate collection of data points from the image of the CO IR spectrum taken earlier. For low resolution spectrums, locate the R-branch and P-branch, use the peak maxima as the data point. If you are unfamiliar on acquiring the data points from a image, follow these steps: Open the website -> Load Image -> Choose File -> Open -> 2D(X-Y) Plot -> Align Axes -> Plot Known $X_{1},X_{2}.Y_{1},Y_{2}$  -> Complete -> Insert Known Values -> Ok -> Add Points -> View Data.

Equation 1: Using the data points from the image, the fundamental frequency can be determined by the following relationship.