Chapter 7, Problem 7.74P

(a)

Interpretation Introduction

Interpretation:

A general expression for the ionization energy of a one electron species is to be written.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{\text{final}}^2}-\frac{1}{n_{\text{initial}}^2}\right)Z^2$ (1)

Answer to Problem 7.74P

The general expression for the ionization energy of one mole of a one electron species is $\Delta E=\left(\text{1}\text{.313014}\times {10}^6\right)Z^2\text{ J/mol}$.

Explanation of Solution

The ionization energy of an atom is the minimum amount of energy required to completely remove the outermost electron from it. An electron is completely removed from an atom when the value of $n_{\text{final}}$ is $\infty$. This is so because the electron is completely free from the influence of the nuclear charge.

Substitute $\infty$ for $n_{\text{final}}$ and 1 for $n_{\text{initial}}$ in equation (1).

$\Delta E=\left(-2.18\times {10}^{-18}\text{ J}\right)\left(\frac{1}{{\infty }^2}-\frac{1}{1^2}\right)Z^2$

For one mole of one electron species, the equation becomes,

$\begin{array}{c}\Delta E=\left(-2.18\times {10}^{-18}\text{ J}\right)\left(\frac{1}{{\infty }^2}-\frac{1}{1^2}\right)Z^2\left(\frac{6.023\times {10}^{23}}{1\text{ mol}}\right)\\ =\left(13.13014\times {10}^5\right)Z^2\text{ J/mol}\\ \text{=}\left(\text{1}\text{.313014}\times {10}^6\right)Z^2\text{ J/mol}\end{array}$

Conclusion

The general expression for the ionization energy of one mole of a one electron species is $\Delta E=\left(\text{1}\text{.313014}\times {10}^6\right)Z^2\text{ J/mol}$.

(b)

Interpretation Introduction

Interpretation:

The ionization energy of ${\text{B}}^{4+}$ is to be calculated.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The general expression for the ionization energy of one mole of a one electron species is $\Delta E=\left(\text{1}\text{.313014}\times {10}^6\right)Z^2\text{ J/mol}$ (2)

Answer to Problem 7.74P

The ionization energy of ${\text{B}}^{4+}$ is $3.28\times {10}^7\text{ J/mol}$.

Explanation of Solution

The symbol $\text{B}$ represents the element boron. The $Z$ value signifies the atomic number of an element. The atomic number of boron is 5.

Substitute 5 for $Z$ in eqution (2).

$\begin{array}{c}\Delta E=\left(\text{1}\text{.313014}\times {10}^6\right){\left(5\right)}^2\text{ J/mol}\\ \text{=}\left(\text{1}\text{.313014}\times {10}^6\right)\left(25\right)\text{ J/mol}\\ =32.82535\times {10}^6\text{ J/mol}\\ =3.28\times {10}^7\text{ J/mol}\end{array}$

Conclusion

The ionization energy of ${\text{B}}^{4+}$ is $3.28\times {10}^7\text{ J/mol}$.

(c)

Interpretation Introduction

Interpretation:

The minimum wavelength required to remove the electron from the $n=3$ level of ${\text{He}}^{+}$ is to be determined.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The equation that relates to the frequency and wavelength of electromagnetic radiation is as follows:

$\nu =\frac{c}{\lambda }$

Here,

$c$ is the speed of light.

$\lambda$ is the wavelength.

$\nu$ is the frequency of the radiation.

Energy is proportional to the frequency and is expressed by the Plank-Einstein equation as follows:

$\Delta E=h\nu$

Here,

$E$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The above relation can be modified as follows:

$\Delta E=h\left(\frac{c}{\lambda }\right)$ (3)

Answer to Problem 7.74P

The minimum wavelength required to remove the electron from the $n=3$ level of ${\text{He}}^{+}$ is $205\text{ nm}$.

Explanation of Solution

Substitute $\infty$ for $n_{\text{final}}$, 3 for $n_{\text{initial}}$ and 2 for $Z$ in equation (1).

$\begin{array}{c}\Delta E=\left(-2.18\times {10}^{-18}\text{ J}\right)\left(\frac{1}{{\infty }^2}-\frac{1}{3^2}\right)\left(2^2\right)\\ =0.9688\times {10}^{-18}\text{ J}\\ =9.688\times {10}^{-19}\text{ J}\end{array}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$, $3\times {10}^8$ for $c$ and $9.688\times {10}^{-19}\text{ J}$ for $\Delta E$ in equation (3).

$9.688\times {10}^{-19}\text{ J}=\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(\frac{3\times {10}^8}{\lambda }\right)$

Rearrange the above equation and calculate the value for $\lambda$ as follows:

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\right)}{9.688\times {10}^{-19}\text{ J}}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =2.053\times {10}^{-7}\text{ m}\\ =205\times {10}^{-9}\text{ m}\end{array}$

Conclusion

The minimum wavelength required to remove the electron from the $n=3$ level of ${\text{He}}^{+}$ is $205\text{ nm}$.

(d)

Interpretation Introduction

Interpretation:

The minimum wavelength required to move the electron from $n=2$ level of ${\text{Be}}^{3+}$, is to be determined.

Concept introduction:

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{\text{final}}^2}-\frac{1}{n_{\text{initial}}^2}\right)Z^2$ (1)

The equation that relates to the frequency and wavelength of electromagnetic radiation is as follows:

$\nu =\frac{c}{\lambda }$

Here,

$c$ is the speed of light.

$\lambda$ is the wavelength.

$\nu$ is the frequency of the radiation.

Energy is proportional to the frequency and is expressed by the Plank-Einstein equation as follows:

$\Delta E=h\nu$

Here,

$E$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The above relation can be modified as follows:

$\Delta E=h\left(\frac{c}{\lambda }\right)$ (3)

Answer to Problem 7.74P

The minimum wavelength required to move the electron from $n=2$ level of ${\text{Be}}^{3+}$, is $22.8\text{ nm}$.

Explanation of Solution

Substitute $\infty$ for $n_{\text{final}}$, 2 for $n_{\text{initial}}$ and 4 for $Z$ in equation (1).

$\begin{array}{c}\Delta E=\left(-2.18\times {10}^{-18}\text{ J}\right)\left(\frac{1}{{\infty }^2}-\frac{1}{2^2}\right)\left(4^2\right)\\ =\left(-2.18\times {10}^{-18}\text{ J}\right)\left(0-\frac{1}{4}\right)\left(16\right)\\ =8.72\times {10}^{-18}\text{ J}\end{array}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$, $3\times {10}^8$ for $c$ and $8.72\times {10}^{-18}\text{ J}$ for $\Delta E$ in equation (3).

$8.72\times {10}^{-18}\text{ J}=\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(\frac{3\times {10}^8}{\lambda }\right)$

Rearrange the above equation and calculate the value for $\lambda$ as follows:

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\right)}{8.72\times {10}^{-18}\text{ J}}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =2.28\times {10}^{-8}\text{ m}\\ =22.8\times {10}^{-9}\text{ m}\end{array}$

Conclusion

The minimum wavelength required to move the electron from $n=2$ level of ${\text{Be}}^{3+}$, is $22.8\text{ nm}$.