Problem 19 - Chemistry and Chemical Reactivity

Question

The energy emitted when an electron moves from a higher energy state to a lower energy state in any atom can be observed as electromagnetic radiation. (a) Which involves the emission of less energy in the \(\mathrm{H}\) atom, an electron moving from \(n=4\) to \(n=2\) or an electron moving from \(n=3\) to \(n=2 ?\) (b) Which involves the emission of more energy in the H atom, an electron moving from \(n=4\) to \(n=1\) or an electron moving from \(n=5\) to \(n=2 ?\) Explain fully.

Step-by-Step Solution

Answer

(a) n=3 to n=2 emits less energy. (b) n=4 to n=1 emits more energy.

Step 1: Understanding the Energy Levels Concept

In any atom, electrons are found in quantized energy levels or orbitals. When an electron transitions between these levels, there is a change in energy which can result in the emission or absorption of light. The energy emitted or absorbed corresponds to a difference in energy between the two levels involved and can be calculated using the formula \[ E = - R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]where \(E\) is the energy, \(R_H\) is the Rydberg constant \(2.18 \times 10^{-18} \text{ J}\), \(n_i\) is the initial energy level, and \(n_f\) is the final energy level.

Step 2: Calculate Energy for Transition n=4 to n=2

For a transition from \(n_i = 4\) to \(n_f = 2\), use the formula:\[ E_{4 \rightarrow 2} = - R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{1}{4} - \frac{1}{16} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{4 - 1}{16} \right) \]\[ = - (2.18 \times 10^{-18}) \times \frac{3}{16} \]\[ = -4.09 \times 10^{-19} \text{ J} \]

Step 3: Calculate Energy for Transition n=3 to n=2

For a transition from \(n_i = 3\) to \(n_f = 2\), use the formula:\[ E_{3 \rightarrow 2} = - R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{1}{4} - \frac{1}{9} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{9 - 4}{36} \right) \]\[ = - (2.18 \times 10^{-18}) \times \frac{5}{36} \]\[ = -3.03 \times 10^{-19} \text{ J} \]

Step 4: Compare Energy Emissions for Part (a)

In comparing 4 to 2 transition and 3 to 2 transition:- Energy \(E_{4 \rightarrow 2} = -4.09 \times 10^{-19} \text{ J}\)- Energy \(E_{3 \rightarrow 2} = -3.03 \times 10^{-19} \text{ J}\)The electron moving from \(n=3\) to \(n=2\) emits less energy than from \(n=4\) to \(n=2\).

Step 5: Calculate Energy for Transition n=4 to n=1

For a transition from \(n_i = 4\) to \(n_f = 1\), use the formula:\[ E_{4 \rightarrow 1} = - R_H \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{1}{1} - \frac{1}{16} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{16 - 1}{16} \right) \]\[ = - (2.18 \times 10^{-18}) \times \frac{15}{16} \]\[ = -2.04375 \times 10^{-18} \text{ J} \]

Step 6: Calculate Energy for Transition n=5 to n=2

For a transition from \(n_i = 5\) to \(n_f = 2\), use the formula:\[ E_{5 \rightarrow 2} = - R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{1}{4} - \frac{1}{25} \right) \]\[ = - (2.18 \times 10^{-18}) \left( \frac{25 - 4}{100} \right) \]\[ = - (2.18 \times 10^{-18}) \times \frac{21}{100} \]\[ = -4.578 \times 10^{-19} \text{ J} \]

Step 7: Compare Energy Emissions for Part (b)

In comparing 4 to 1 transition and 5 to 2 transition:- Energy \(E_{4 \rightarrow 1} = -2.04375 \times 10^{-18} \text{ J} \)- Energy \(E_{5 \rightarrow 2} = -4.578 \times 10^{-19} \text{ J} \)The electron moving from \(n=4\) to \(n=1\) emits more energy than from \(n=5\) to \(n=2\).

Key Concepts

Energy Levels

In an atom, electrons occupy specific regions called energy levels or shells. These levels are quantized, meaning they have specific energy values that electrons can possess.

The different energy levels correspond to different integers, denoted by the principal quantum number, \(n\).

An electron in a higher energy level has more energy compared to an electron in a lower energy level.

• The first energy level (\(n=1\)) is the lowest and closest to the nucleus, having the least energy.
• As \(n\) increases, the energy level moves further from the nucleus and has higher energy.

When electrons transition between these energy levels, they emit or absorb energy, making these processes essential to understanding atomic emissions and absorptions.

This emitted or absorbed energy can be calculated using various formulas, leading us to the next important concept, the Rydberg constant.

Rydberg Constant

The Rydberg constant, denoted as \(R_H\), is a fundamental physical constant that helps calculate the energy associated with electron transitions in hydrogen-like atoms. It serves as a crucial part of quantifying how much energy is emitted or absorbed when an electron changes its energy level.

For hydrogen, the Rydberg constant has a value of \(2.18 \times 10^{-18}\) Joules. This constant is pivotal in determining the energy differences during electron transitions, especially in hydrogen atoms.

• It forms part of the formula used to determine the energy change when electrons move between different shells: \[ E = - R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
• The formula shows that the energy change is proportional to the inverse of the square of the principal quantum numbers \(n\).

Understanding the Rydberg constant enhances our ability to predict and explain the behavior of atoms and electrons concerning energy emission and absorption.

Electron Transitions

Electron transitions occur when an electron moves from one energy level to another within an atom. This movement comes with the exchange of energy, either releasing or absorbing it.

These transitions are often visible as spectral lines in atomic emission or absorption spectra. The energy level from which an electron departs and the level to which it moves determine the energy exchange:

• Moving from a higher \(n\) to a lower \(n\) releases energy, resulting in emission.
• Moving from a lower \(n\) to a higher \(n\) absorbs energy, resulting in absorption.

The energy associated with each transition is dependent on the difference between the two energy levels involved. The values of \(n_i\) and \(n_f\), initial and final energy levels, directly affect this energy difference.

For instance, electron transitions in hydrogen between levels \(n=4\) and \(n=2\), or \(n=3\) and \(n=2\), can be calculated to show varying energies being released, each corresponding to different quantities of emitted light.

Electromagnetic Radiation

When electrons undergo transitions between different energy levels, they emit or absorb energy in the form of electromagnetic radiation. This radiation can be in the form of visible light, ultraviolet light, or other types of electromagnetic waves, depending on the energy involved in the transition.

The characteristics of electromagnetic radiation emitted or absorbed depend on the difference in energy levels, which dictates the wavelength and frequency of the light:

• Emitted electromagnetic radiation has a wavelength inversely proportional to the energy change \(E\).
• Higher energy transitions correspond to shorter wavelengths and higher frequencies.

Electromagnetic radiation is a key observable feature in many scientific fields, helping scientists analyze and understand atomic structures.

By studying these emissions in various atoms, we can gain insights into an atom's characteristics and the transitions occurring within it.