Chapter 12, Problem 12.15E

(a) Assume that the electronic energy of $\text{Li}$ was a product of three hydrogen-like wavefunctions with principal quantum number equal to $1$. What would be the total energy of $\text{Li}$?

(b) Assume that two of the principal quantum numbers are $1$ and the third principal quantum number is $2$. Calculate the estimated electronic energy.

(c) Compare both energies with an experimental value of $-3.26\times {10}^{-17}\text{J}$. Which estimate is better? Is there any reason you might assume that this estimate would be better from the start?

Interpretation Introduction

(a)

Interpretation:

The total energy of $\text{Li}$ is to be calculated by using given assumption.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time.The time-independent Schrödinger equation is represented as,

$\left[-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+\widehat{V}\left(x\right)\right]\Psi =E\Psi$

Where,

• $m$ represents the mass of the particle.

• $x$ represents the coordinate of the particle at the x-axis.

• $\widehat{V}$ represents the potential energy operator.

• $\Psi$ represents the wavefunction.

• $E$ represents the energy of the particle.

• $\hslash$ represents a constant.

Answer to Problem 12.15E

The total energy of $\text{Li}$ is $-5.8833\times {\text{10}}^{-17}\text{J}$.

Explanation of Solution

The atomic number of lithium is $3$.

The given principal quantum number is $1$.

The charge on an electron is $1.602\times {10}^{-19}\text{C}$.

The reduced mass of the electron is $9.109\times {10}^{-31}\text{kg}$.

It is assumed that the electronic energy of $\text{Li}$ was a product of three hydrogen-like wavefunctions. It is mathematically represented as,

${\Psi }_{\text{Li}}={\Psi }_{\text{H, 1}}\times {\Psi }_{\text{H, 2}}\times {\Psi }_{\text{H, 3}}$

Therefore, the Schrödinger equation for lithium is represented as,

$\left[\begin{array}{l}\left(-\frac{{\hslash }^2}{2\mu }{\nabla }_1^2-\frac{3e^2}{4\pi {\in }_0{r}_1}\right)+\left(-\frac{{\hslash }^2}{2\mu }{\nabla }_2^2-\frac{3e^2}{4\pi {\in }_0{r}_2}\right)\\ +\left(-\frac{{\hslash }^2}{2\mu }{\nabla }_3^2-\frac{3e^2}{4\pi {\in }_0{r}_3}\right)\end{array}\right]{\Psi }_{\text{H, 1}}{\Psi }_{\text{H, 2}}{\Psi }_{\text{H, 3}}\approx E_{\text{Li}}{\Psi }_{\text{H, 1}}{\Psi }_{\text{H, 2}}{\Psi }_{\text{H, 3}}$

The above equation can be written as individual energies of each wave function as,

$\begin{array}{l}{E}_{\text{Li}}{\Psi }_{\text{H, 1}}=-\frac{{\hslash }^2}{2\mu }{\nabla }_1^2-\frac{3e^2}{4\pi {\in }_0{r}_1}\\ E_{\text{Li}}{\Psi }_{\text{H, 2}}=-\frac{{\hslash }^2}{2\mu }{\nabla }_2^2-\frac{3e^2}{4\pi {\in }_0{r}_2}\\ E_{\text{Li}}{\Psi }_{\text{H, 3}}=-\frac{{\hslash }^2}{2\mu }{\nabla }_3^2-\frac{3e^2}{4\pi {\in }_0{r}_3}\end{array}$

Therefore, the energy of the lithium atom can be represented as the sum of the eigenvalue of all three energies. …(1)

The eigenvalue of energy is represented as,

$E=-\frac{Z^2{e}^4\mu }{8{\in }_0^2{h}^2{n}^2}$ …(2)

Where,

• $Z$ represents the atomic number of the atom.

• $e$ represents the charge on an electron.

• $\mu$ represents the reduced mass of the electron.

• ${\in }_0$ represents a constant with a value $8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}$

• $h$ represents Plank’s constant $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

• $n$ represents principal quantum number.

Substitute the value of $Z$, $e$, $\mu$, ${\in }_0$, $h$ and $n$ in the equation (2).

$\begin{array}{rll}{E}_1& =& -\frac{{\left(3\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^4\left(9.109\times {10}^{-31}\text{kg}\right)}{8{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)}^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2{\left(1\right)}^2}\\ & =& -\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\end{array}$

The values of all three energies will be same as the values of principal quantum number for all three are same.

Substitute the value of energies in the equation (1).

$\begin{array}{rll}E& =& -\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}+\left(-\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\right)+\left(-\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\right)\\ & =& -5.8833\times {\text{10}}^{-17}\text{J}\end{array}$

Therefore, the total energy of $\text{Li}$ is $-5.8833\times {\text{10}}^{-17}\text{J}$.

Conclusion

The total energy of $\text{Li}$ is $-5.8833\times {\text{10}}^{-17}\text{J}$.

Interpretation Introduction

(b)

Interpretation:

The estimated electronic energy is to be calculated by using given assumption.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time.The time-independent Schrödinger equation is represented as,

$\left[-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+\widehat{V}\left(x\right)\right]\Psi =E\Psi$

Where,

• $m$ represents the mass of the particle.

• $x$ represents the coordinate of the particle at the x-axis.

• $\widehat{V}$ represents the potential energy operator.

• $\Psi$ represents the wavefunction.

• $E$ represents the energy of the particle.

• $\hslash$ represents a constant.

Answer to Problem 12.15E

The estimated electronic energy of $\text{Li}$ is $-4.4125\times {\text{10}}^{-17}\text{J}$.

Explanation of Solution

The atomic number of lithium is $3$.

The given principal quantum number for first two energies is $1$.

The given principal quantum number for Third energies is $2$.

The charge on an electron is $1.602\times {10}^{-19}\text{C}$.

The reduced mass of the electron is $9.109\times {10}^{-31}\text{kg}$.

The energy of the lithium atom can be represented as the sum of the eigenvalue of all three energies.

$E=E_{\text{1}}+E_2+E_3$ …(1)

The eigenvalue of energy is represented as,

$E=-\frac{Z^2{e}^4\mu }{8{\in }_0^2{h}^2{n}^2}$ …(2)

Where,

• $Z$ represents the atomic number of the atom.

• $e$ represents the charge on an electron.

• $\mu$ represents the reduced mass of the electron.

• ${\in }_0$ represents a constant with a value $8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}$

• $h$ represents Plank’s constant $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

• $n$ represents principal quantum number.

Substitute the value of $Z$, $e$, $\mu$, ${\in }_0$, $h$ and $n=1$ in the equation (2).

$\begin{array}{rll}{E}_1& =& -\frac{{\left(3\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^4\left(9.109\times {10}^{-31}\text{kg}\right)}{8{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)}^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2{\left(1\right)}^2}\\ & =& -\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\end{array}$

The values first two of energies will be same as the values of principal quantum number for these two are same.

Substitute the value of $Z$, $e$, $\mu$, ${\in }_0$, $h$ and $n=2$ in the equation (2).

$\begin{array}{rll}{E}_3& =& -\frac{{\left(3\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^4\left(9.109\times {10}^{-31}\text{kg}\right)}{8{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)}^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2{\left(2\right)}^2}\\ & =& -\text{0}\text{.4903}\times {\text{10}}^{-17}\text{J}\end{array}$

Substitute the value of energies in the equation (1).

$\begin{array}{rll}E& =& \left(-\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\right)+\left(-\text{1}\text{.9611}\times {\text{10}}^{-17}\text{J}\right)+\left(-\text{0}\text{.4903}\times {\text{10}}^{-17}\text{J}\right)\\ & =& -4.4125\times {\text{10}}^{-17}\text{J}\end{array}$

Therefore, the estimated electronic energy of $\text{Li}$ is $-4.4125\times {\text{10}}^{-17}\text{J}$.

Conclusion

The estimated electronic energy of $\text{Li}$ is $-4.4125\times {\text{10}}^{-17}\text{J}$.

Interpretation Introduction

(c)

Interpretation:

The corresponding estimated energy is to be compared with the experimental energy. The estimate that would be better is to be identified. Whether there is any reason which might assume that the corresponding estimate would better from the start or not is to be stated.

Concept introduction:

Experimental value is the value that is obtained by physically performing the experiment. Calculated value is obtained by mathematically calculating the expected value of quantity. The difference between calculated value and experimental value gives an idea about the efficiency of the method used.

Answer to Problem 12.15E

The second calculated value is more close to the experimental value therefore, the better value of energy is $-4.4125\times {\text{10}}^{-17}\text{J}$. Yes, there is a reason which might assume that the second estimate would be better from the start.

Explanation of Solution

The experimental value of energy of lithium atom is $-3.26\times {10}^{-17}\text{J}$.

The total energy of $\text{Li}$ is $-5.8833\times {\text{10}}^{-17}\text{J}$.

The estimated electronic energy of $\text{Li}$ is $-4.4125\times {\text{10}}^{-17}\text{J}$.

The second calculated value is more close to the experimental value. Therefore, the second value would be better.

For $n=1$, only $s$ orbital is present and only two electrons can occupy this orbital. The first assumption assumes that all three electrons have $n=1$ which is incorrect. Therefore, there is a reason which might assume that the second estimate would be better from the start.

Conclusion

The second calculated value is more close to the experimental value therefore, the better value of energy is $-4.4125\times {\text{10}}^{-17}\text{J}$. Yes, there is a reason which might assume that the second estimate would be better from the start.