Chapter 8, Problem 8.2IA

(a)

Interpretation Introduction

Interpretation:

The relation for the separation of energy for hydrogen atoms with large value of $n$ has to be derived.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons.  Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom.  The electrons are arranged in the orbit of atoms according to their energy.

Answer to Problem 8.2IA

The relation for the separation of energy for hydrogen atoms with large value of $n$ has been derived.

Explanation of Solution

The energy level for an atom is given by the expression shown below.

  $E_n=-\frac{hc{R}_{\text{H}}}{n^2}$        (1)

Where,

• $h$ is Plank’s constant with a value of $6.62608\times {10}^{-34}\text{J s}$.
• $c$ is the speed of light with the value of $2.997945\times {10}^8\text{m}{\text{s}}^{-1}$.
• $n$ is the energy level.
• $R_{\text{H}}$ Rydberg constant of the hydrogen atom.

The difference in energy of two energy levels can be given by the expression as shown below.

    $\Delta E_{n\to n+1}=E_{n+1}-E{n}_{}$

Where,

• $E_{n+1}$ is the energy of higher level.
• $E{n}_{}$ is the energy of lower level.
• $n$ is the principal quantum number of energy level.

Substitute the value of $E_{n+1}$ and $E_n$ in the above equation.

  $\begin{array}{c}\Delta E_{n\to n+1}=-\frac{hc{R}_{\text{H}}}{{\left(n+1\right)}^2}-\left(-\frac{hc{R}_{\text{H}}}{n^2}\right)\\ =-hc{R}_{\text{H}}\left(\frac{1}{{\left(n+1\right)}^2}-\frac{1}{n^2}\right)\end{array}$

Therefore, the relation for the separation of energy for hydrogen atoms with large value of $n$ is shown below.

    $\Delta E_{n\to n+1}=-hc{R}_{\text{H}}\left(\frac{1}{{\left(n+1\right)}^2}-\frac{1}{n^2}\right)$

(b)

Interpretation Introduction

Interpretation:

The value of energy separation, the average radius and the ionization energy for $n=100$ have to be calculated.

Concept introduction:

The energy required to remove an electron from an ion or isolated atom is known as its ionization energy.  The ionization energy of metals are generally lower than the ionization energy of nonmetals.  The formula of ionization energy for an atom is given by the expression shown below.

  $I=\frac{hc{R}_{\text{H}}}{n^2}$

Answer to Problem 8.2IA

The value of energy separation, the average radius and the ionization energy for $n=100$ are $\underset{\_}{4.2926\times 10^{-24}J}$, $\underset{\_}{7.935\times 10^{-7}m}$ and $\underset{\_}{2.179\times 10^{-22}J}$ respectively.

Explanation of Solution

The value of $n$ for energy level is $100$.

The atomic number of hydrogen is $1$.

The quantum number $\left(l,m_l\right)$ for hydrogen atom is $\left(0,0\right)$.

The relation for the separation of energy for hydrogen atoms with large value of $n$ is shown below.

    $\Delta E_{n\to n+1}=-hc{R}_{\text{H}}\left(\frac{1}{{\left(n+1\right)}^2}-\frac{1}{n^2}\right)$

Where,

• $h$ is Plank’s constant with a value of $6.62608\times {10}^{-34}\text{J s}$.
• $c$ is the speed of light with the value of $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.
• $n$ is the energy level.
• $R_{\text{H}}$ Rydberg constant of the hydrogen atom $\left(109677{\text{cm}}^{-1}\right)$.

Substitute the value of $h$, $c$, $R_{\text{H}}$ and $n$ in the above equation.

    $\begin{array}{c}\Delta E_{n\to n+1}=\left(\begin{array}{l}-\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(109677{\text{cm}}^{-1}\right)\left(\frac{1}{{\left(100+1\right)}^2}-\frac{1}{{\left(100\right)}^2}\right)\end{array}\right)\\ =-\left(2.179\times {10}^{-18}\right)\left(\frac{1}{10201}-\frac{1}{10000}\right)\text{J}\\ =-\left(2.179\times {10}^{-18}\right)\left(-1.97\times {10}^{-6}\right)\text{J}\\ =\underset{\_}{4.2926\times 10^{-24}J}\end{array}$

The formula for average radius of hydrogenic orbital is given by the expression as shown below.

    $r_{n,l,m_l}=\frac{n^2{a}_0}{Z}\left\{1+\frac{1}{2}\left[1-\frac{l\left(l+1\right)}{n^2}\right]\right\}$

Where,

• $a_0$ is the Bohr’s radius $\left(5.29\times {10}^{-11}\text{m}\right)$.
• $Z$ is the atomic number of element.
• $l$ is the azimuthal quantum number of angular momentum.
• $m_l$ is the magnetic quantum number.

Substitute the values of $a_0$, $Z$, $l$, $m_l$, and $n$ in the above equation.

  $\begin{array}{c}{r}_{100,0,0}=\frac{{\left(100\right)}^2\left(5.29\times {10}^{-11}\text{m}\right)}{\left(1\right)}\left\{1+\frac{1}{2}\left[1-\frac{\left(0\right)\left(0+1\right)}{{\left(100\right)}^2}\right]\right\}\\ =\left(5.29\times {10}^{-7}\text{m}\right)\left(1+\frac{1}{2}\right)\\ =\left(5.29\times {10}^{-7}\text{m}\right)\left(\frac{3}{2}\right)\\ =\underset{\_}{7.935\times 10^{-7}m}\end{array}$

The ionization energy of the atom is given by the expression as shown below.

    $I=\frac{hc{R}_{\text{H}}}{n^2}$

Substitute the values of $h$, $c$, $R_{\text{H}}$ and $n$ in the above equation.

    $\begin{array}{c}I=\frac{\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(109677{\text{cm}}^{-1}\right)}{{\left(100\right)}^2}\\ =\frac{\left(2.179\times {10}^{-18}\right)}{10000}\text{ J}\\ =\underset{\_}{2.179\times 10^{-22}J}\end{array}$

Therefore, the value of energy separation, the average radius and the ionization energy for $n=100$ are $\underset{\_}{4.2926\times 10^{-24}J}$, $\underset{\_}{7.935\times 10^{-7}m}$ and $\underset{\_}{2.179\times 10^{-22}J}$ respectively.

(c)

Interpretation Introduction

Interpretation:

Whether a thermal collision with another hydrogen atom would ionize the Rydberg atom with $n=100$ or not has to be predicted.

Concept introduction:

As mentioned in the concept introduction in part (b).

Answer to Problem 8.2IA

A thermal collision with another hydrogen atom will ionize the Rydberg atom with $n=100$.

Explanation of Solution

The ionization energy of hydrogen atom with $n=100$ is $2.179\times {10}^{-22}\text{J}$.

The standard room temperature is $298\text{K}$.

The formula of thermal energy is given by the expression as shown below.

  $E=kT$

Where,

• $k$ is Boltzmann constant $\left(1.3806488\times {10}^{-23}\text{ J}{\text{K}}^{-1}\right)$.
• $T$ is the temperature.

Substitute the values of $k$, and $T$ in the above equation.

    $\begin{array}{c}E=\left(1.3806488\times {10}^{-23}\text{ J}{\text{K}}^{-1}\right)\left(298\text{K}\right)\\ =4.1143\times {10}^{-21}\text{ J}\end{array}$

The thermal energy provided by collision is more than the energy required to ionize atom. Therefore, a thermal collision with another hydrogen atom will ionize the Rydberg atom with $n=100$.

(d)

Interpretation Introduction

Interpretation:

The minimum velocity of the second atom is required to ionize the Rydberg atom with $n=100$ is to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (b).

Answer to Problem 8.2IA

The minimum velocity of the second atom is required to ionize the Rydberg atom with $n=100$ is $\underset{\_}{509.9m{s}^{-1}}$.

Explanation of Solution

The ionization energy of hydrogen atom with $n=100$ is $2.179\times {10}^{-22}\text{J}$.

The mass of hydrogen atom is $1.008\text{u}$.

The mass of hydrogen atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{H}}=\left(1.008\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =1.6738\times {10}^{-27}\text{kg}\end{array}$

The kinetic energy of hydrogen atom can be given by the expression as shown below.

    $E_{\text{k}}=\frac{1}{2}{m}_{\text{H}}{v}^2$        (2)

Where,

• $m_{\text{H}}$ is the mass of the hydrogen atom.
• $v$ is the velocity of hydrogen atom.

The kinetic energy of hydrogen atom must be greater than or equal to ionization energy hydrogen atom.

Therefore, minimum value of kinetic energy of hydrogen atom is $2.179\times {10}^{-22}\text{J}$.

Substitute the values of $m_{\text{H}}$, and $E_{\text{k}}$ in the equation (2).

  $\begin{array}{c}\left(2.179\times {10}^{-22}\text{J}\right)=\frac{1}{2}\left(1.6738\times {10}^{-27}\text{kg}\right)v^2\\ \left(2.179\times {10}^{-22}\text{J}\right)=\left(0.8369\times {10}^{-27}\text{kg}\right)v^2\end{array}$

Rearrange the above expression for the value of $v$.

  $\begin{array}{c}v={\left(\frac{\left(2.179\times {10}^{-22}\text{J}\right)\left(\frac{1{\text{ kg m}}^2{\text{ s}}^{-2}}{1\text{ J}}\right)}{\left(0.8369\times {10}^{-27}\text{kg}\right)}\right)}^{1/2}\\ ={\left(2.60\times {10}^5\right)}^{1/2}{\text{ m s}}^{-1}\\ =\underset{\_}{509.9m{s}^{-1}}\end{array}$

Therefore, the minimum velocity of the second atom is required to ionize the Rydberg atom with $n=100$ is $\underset{\_}{509.9m{s}^{-1}}$.

(e)

Interpretation Introduction

Interpretation:

The radial wavefunction for a $100\text{s}$ orbital has to be depicted.

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 square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 8.2IA

The radial wavefunction for a $100\text{s}$ orbital will have $99$. The radial wavefunction for a $100\text{s}$ orbital proportional to the product of ${\rho }^{n-1}$ and $e^{-\rho /2n}$. The graph will have a similar shape as $1\text{s}$, $2\text{s}$ and $\text{3s}$ orbitals. The expected radial wavefunction for a $100\text{s}$ orbital is shown below.

Explanation of Solution

The graphs of radial wavefunction of $1\text{s}$, $2\text{s}$ and $\text{3s}$ orbitals is shown below.

Figure 1

For $100\text{s}$ orbital, the number of nodes will be $99$.

The radial wavefunction of $100\text{s}$ will be similar as $1\text{s}$, $2\text{s}$ and $\text{3s}$ orbitals.  It is proportional to the product of ${\rho }^{n-1}$ and $e^{-\rho /2n}$.  The coefficient $\rho$ depends inversely on the value of radius.  The expected radial wavefunction of $100\text{s}$ is shown below.

    $\begin{array}{c}R\left(r\right)\propto {\rho }^{100-1}{e}^{-\rho /2\left(100\right)}\\ \propto {\rho }^{99}{e}^{-\rho /200}\end{array}$

Therefore, the expected graph of radial wavefunction for a $100\text{s}$ orbital is shown below.

Figure 2