Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value ${\text{ψ}}^{\text{2}}$, ${\text{ψ}}^{\text{2}}$ and $\text{4}\pi r^2{\text{ψ}}^{\text{2}}$ for different value of $r$ are to be calculated. Also, the graph between $\text{ψ}$ versus radius, ${\text{ψ}}^{\text{2}}$ versus radius and ${\text{4πr}}^{\text{2}}{\text{ψ}}^{\text{2}}$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

  $\stackrel{\wedge }{H}\text{ψ}=E\text{ψ}$

Here,

$E$ is the energy of the atom.

$\stackrel{\wedge }{H}$ is the Hamiltonian operator.

$\text{ψ}$ is the wave function.

In the Schrodinger wave equation, $\text{ψ}$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $\left({\text{ψ}}^2\right)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol ${\text{ψ}}^2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

  $\text{ψ}=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_o}\right)}^{\frac{3}{2}}{e}^{\frac{-r}{a_o}}$        (1)

Here,

$\text{ψ}$ is the wave function.

$r$ is the distance from the nucleus.

Answer to Problem 7.79P

The value of ${\text{ψ}}^{\text{2}}$, ${\text{ψ}}^{\text{2}}$ and $\text{4}\pi r^2{\text{ψ}}^{\text{2}}$ for different value of $r$ are as follows:

$\begin{array}{cccc}r\left(\text{pm}\right)& \text{ψ}& {\text{ψ}}^{\text{2}}& \text{4}\pi r^2{\text{ψ}}^{\text{2}}\\ 0& 1.47\times {10}^{-3}& 2.15\times {10}^{-6}& 0\\ 50& 0.570\times {10}^{-3}& 0.325\times {10}^{-6}& 1.02\times {10}^{-2}\\ 100& 0.221\times {10}^{-3}& 0.0491\times {10}^{-6}& 0.0616\times {10}^{-2}\\ 200& 0.0335\times {10}^{-3}& 0.00112\times {10}^{-6}& 0.0563\times {10}^{-2}\end{array}$

The graph between $\text{ψ}$ versus radius, ${\text{ψ}}^{\text{2}}$ versus radius and ${\text{4πr}}^{\text{2}}{\text{ψ}}^{\text{2}}$ versus r is as follows:

Explanation of Solution

Substitute $52.92\text{ pm}$ for $a_o$, $0$ for $r$ in the equation (1).

  $\begin{array}{c}\text{ψ}=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{52.92\text{ pm}}\right)}^{\frac{3}{2}}{e}^{\frac{-0}{52.92\text{ pm}}}\\ =1.465532\times {10}^{-3}\\ =1.47\times {10}^{-3}\end{array}$

The expression to calculate the probability density of electrons is as follows:

  $\text{Probability density of electron}={\text{ψ}}^{\text{2}}$        (2)

Substitute $1.465532\times {10}^{-3}$ for $\text{ψ}$ in the equation (2).

  $\begin{array}{c}\text{Probability density of electron}={\left(1.465532\times {10}^{-3}\right)}^{\text{2}}\\ =2.15\times {10}^{-6}{\text{ pm}}^{-3}\end{array}$

The expression to calculate the mass of an electron is as follows:

  $\text{Mass of electron}={\text{4πr}}^{\text{2}}{\text{ψ}}^{\text{2}}$        (3)

Substitute $2.15\times {10}^{-6}{\text{ pm}}^{-3}$ for ${\text{ψ}}^{\text{2}}$ and $0$ for $r$ in the equation (3).

  $\begin{array}{c}\text{Mass of electron}=\text{4π}{\left(0\right)}^{\text{2}}\left(2.15\times {10}^{-6}{\text{ pm}}^{-3}\right)\\ =0\end{array}$

Substitute $52.92\text{ pm}$ for $a_o$, $50\text{ pm}$ for $r$ in the equation (1).

  $\begin{array}{c}\text{ψ}=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{52.92\text{ pm}}\right)}^{\frac{3}{2}}{e}^{\frac{-50\text{pm}}{52.92\text{ pm}}}\\ =1.465532\left(e^{\frac{-50\text{pm}}{52.92\text{ pm}}}\right)\\ =5.69724\times {10}^{-4}\end{array}$

Substitute $5.69724\times {10}^{-4}$ for $\text{ψ}$ in the equation (2).

  $\begin{array}{c}\text{Probability density of electron}={\left(5.69724\times {10}^{-4}\right)}^{\text{2}}\\ =3.24585\times {10}^{-7}{\text{ pm}}^{-3}\end{array}$

Substitute $3.24585\times {10}^{-7}$ for ${\text{ψ}}^{\text{2}}$ and $50\text{ pm}$ for $r$ in the equation (3).

  $\begin{array}{c}\text{Mass of electron}=\text{4π}{\left(50\text{ pm}\right)}^{\text{2}}\left(3.24585\times {10}^{-7}\right)\\ =1.0197\times {10}^{-2}{\text{pm}}^{-1}\end{array}$

Substitute $52.92\text{ pm}$ for $a_o$, $100\text{ pm}$ for $r$ in the equation (1).

  $\begin{array}{c}\text{ψ}=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{52.92\text{ pm}}\right)}^{\frac{3}{2}}{e}^{\frac{-100\text{pm}}{52.92\text{ pm}}}\\ =1.465532\left(e^{\frac{-50\text{pm}}{52.92\text{ pm}}}\right)\\ =0.221\times {10}^{-3}\end{array}$

Substitute $0.221\times {10}^{-3}$ for $\text{ψ}$ in the equation (2).

  $\begin{array}{c}\text{Probability density of electron}={\left(0.221\times {10}^{-3}\right)}^{\text{2}}\\ =0.0491\times {10}^{-6}{\text{ pm}}^{-3}\end{array}$

Substitute $0.0491\times {10}^{-6}{\text{ pm}}^{-3}$ for ${\text{ψ}}^{\text{2}}$ and $50\text{ pm}$ for $r$ in the equation (3).

  $\begin{array}{c}\text{Mass of electron}=\text{4π}{\left(50\text{ pm}\right)}^{\text{2}}\left(0.0491\times {10}^{-6}{\text{ pm}}^{-3}\right)\\ =0.616\times {10}^{-2}{\text{pm}}^{-1}\end{array}$

Substitute $52.92\text{ pm}$ for $a_o$, $200\text{ pm}$ for $r$ in the equation (1).

  $\begin{array}{c}\text{ψ}=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{52.92\text{ pm}}\right)}^{\frac{3}{2}}{e}^{\frac{-200\text{pm}}{52.92\text{ pm}}}\\ =1.465532\left(e^{\frac{-50\text{pm}}{52.92\text{ pm}}}\right)\\ =0.0335\times {10}^{-3}\end{array}$

Substitute $0.0335\times {10}^{-3}$ for $\text{ψ}$ in the equation (2).

  $\begin{array}{c}\text{Probability density of electron}={\left(0.0335\times {10}^{-3}\right)}^{\text{2}}\\ =0.00112\times {10}^{-6}{\text{ pm}}^{-3}\end{array}$

Substitute $0.00112\times {10}^{-6}{\text{ pm}}^{-3}$ for ${\text{ψ}}^{\text{2}}$ and $200\text{ pm}$ for $r$ in the equation (3).

  $\begin{array}{c}\text{Mass of electron}=\text{4π}{\left(50\text{ pm}\right)}^{\text{2}}\left(0.00112\times {10}^{-6}{\text{ pm}}^{-3}\right)\\ =0.0563\times {10}^{-2}{\text{ pm}}^{-1}.\end{array}$

The graph between $\text{ψ}$ versus radius, ${\text{ψ}}^{\text{2}}$ versus radius and ${\text{4πr}}^{\text{2}}{\text{ψ}}^{\text{2}}$ versus r is as follows:

Conclusion

The graph between ${\text{ψ}}^{\text{2}}$ versus radius plot has zero nodes.