Chapter 7, Problem 7.24QP

Investigating Energy Levels

Consider the hypothetical atom X that has one electron like the H atom but has different energy levels. The energies of an electron in an X atom are described by the equation

$E=-\frac{R_{\text{H}}}{n^3}$

where R_H is the same as for hydrogen (2.179 × 10⁻¹⁸ J). Answer the following questions, without calculating energy values.

1. a How would the ground-state energy levels of X and H compare?
2. b Would the energy of an electron in the n = 2 level of H be higher or lower than that of an electron in the n = 2 level of X? Explain your answer.
3. c How do the spacings of the energy levels of X and H compare?
4. d Which would involve the emission of a higher frequency of light, the transition of an electron in an H atom from the n = 5 to the n = 3 level or a similar transition in an X atom?
5. e Which atom, X or H, would require more energy to completely remove its electron?
6. f A photon corresponding to a particular frequency of blue light produces a transition from the n = 2 to the n = 5 level of a hydrogen atom. Could this photon produce the same transition (n = 12 to n = 5) in an atom of X? Explain.

Interpretation Introduction

Interpretation:

The energy of ground state of atom X and H has to be compared.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Explanation of Solution

To compare: The energy of ground state of atom X and H.

The energy of ground state of atom X and H are same because the n value of ground state is same.

Conclusion

The energy of ground state of atom X and H was compared.

Interpretation Introduction

Interpretation:

The energy of atom X in level two and H in level two has to be compared.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Answer to Problem 7.24QP

Energy of an electron in X atom is lower than H atom.

Explanation of Solution

To compare: The energy of atom X in level two and H in level two.

When n value of H atom is two then $\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}$,  becomes $\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{2^{\text{2}}}$.  And the n value of X atom becomes $\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{2^3}$ which is less compared to H atom.  Thus, energy of an electron in X atom is lower than H atom.

Conclusion

By using the formula of energy levels, the energy of atom X in level two and H in level two was compared.

Interpretation Introduction

Interpretation:

The energy level spacing of atom X and H has to be compared.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Explanation of Solution

To compare: The energy level spacing of atom X and H.

The energy level spacing of hydrogen is greater than X because the $\frac{\text{1}}{{\text{n}}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{(n+1)}}^{\text{2}}}$ is more than $\frac{\text{1}}{{\text{n}}^3}\text{-}\frac{\text{1}}{{\text{(n+1)}}^3}$.

The ratio of energy of H atom and X atom is

  $\begin{array}{l}\frac{{\text{E}}_{\text{H}}}{{\text{E}}_{\text{x}}}\text{=}\frac{\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}}{\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{3}}}}\text{=}\text{n}\\ \\ {\text{E}}_{\text{H}}{\text{ = nE}}_{\text{X}}\text{.}\end{array}$

Conclusion

The energy level spacing of atom X and H was compared using the formula for energy levels.

Interpretation Introduction

Interpretation:

The atom which emits higher frequency of light when transition occurs has to be identified.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Answer to Problem 7.24QP

Hydrogen emits more frequency of light when transition occurs.

Explanation of Solution

To identify: The atom which emits higher frequency of light when transition occurs.

Energy and frequency are directly proportional to each other.  Moreover the energy level spacing of hydrogen is greater than X atom.  Thus, hydrogen emits more frequency of light when transition occurs.

Conclusion

The atom which emits higher frequency of light when transition occurs was identified.

Interpretation Introduction

Interpretation:

The atom which needs more energy to remove electron has to be identified.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Explanation of Solution

To identify: The atom which needs more energy to remove electron.

When the electrons in both atom present in ground state, then same amount of energy is utilized to remove electron.  But, it requires different energies when electron is in different state.

Conclusion

The atom which needs more energy to remove an electron was identified.

Interpretation Introduction

Interpretation:

The frequency of light used for transition from $\text{n}\text{=}\text{2}\text{to}\text{n}\text{=}\text{5}$ level for both atoms are same or not has to be explained.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Explanation of Solution

To explain: The frequency of light used for transition from $\text{n}\text{=}\text{2}\text{to}\text{n}\text{=}\text{5}$ level for both atoms are same or not.

Because of different energy spacing of both atoms, the required energy for transition from $\text{n}\text{=}\text{2}\text{to}\text{n}\text{=}\text{5}$ differs.  Therefore, both atoms requires different frequency of light.

Conclusion

The frequency of light used for transition from $\text{n}\text{=}\text{2}\text{to}\text{n}\text{=}\text{5}$ level for both atoms are same or not was explained.