Problem 20 - Chemistry and Chemical Reactivity

Question

If energy is absorbed by a hydrogen atom in its ground state, the atom is excited to a higher energy state. For example, the excitation of an electron from \(n=1\) to \(n=3\) requires radiation with a wavelength of \(102.6 \mathrm{nm}\) Which of the following transitions would require radiation of longer wavelength than this? (a) \(n=2\) to \(n=4\) (b) \(n=1\) to \(n=4\) (c) \(n=1\) to \(n=5\) (d) \(n=3\) to \(n=5\)

Step-by-Step Solution

Answer

Transitions n=2 to n=4 and n=3 to n=5 require longer wavelengths.

Step 1: Understanding Energy and Wavelength

In this problem, we are dealing with transitions between different energy levels of a hydrogen atom. The energy difference between two energy levels determines the wavelength of radiation required for transition. Since energy and wavelength are inversely related (according to the equation \( E = \frac{hc}{\lambda} \)), higher energy transitions correspond to shorter wavelengths and vice versa.

Step 2: Energy Level Transitions

The energy levels of a hydrogen atom are quantized and calculated using the formula: \( E_n = -\frac{13.6}{n^2} \text{ eV} \), where \( n \) is the principal quantum number. To compare wavelengths for different transitions, calculate the energy differences \( \Delta E \) for each transition provided.

Step 3: Calculate Energy Difference for n=1 to n=3

The energy difference from \( n=1 \) to \( n=3 \) is given as corresponding to a wavelength of \( 102.6 \text{ nm} \). Calculate \( E_3 - E_1 \) as follows: \[ \Delta E_{1-3} = E_3 - E_1 = -\frac{13.6}{3^2} - \left(-\frac{13.6}{1^2} \right) = 12.09 \text{ eV} \]

Step 4: Calculate Energy Difference for Other Transitions

Calculate the energy difference for each option: - **(a)** \( n=2 \) to \( n=4 \): \[ \Delta E_{2-4} = E_4 - E_2 = -\frac{13.6}{4^2} - -\frac{13.6}{2^2} = -0.85 \text{ eV} \] - **(b)** \( n=1 \) to \( n=4 \): \[ \Delta E_{1-4} = E_4 - E_1 = -\frac{13.6}{4^2} - -\frac{13.6}{1^2} = 12.75 \text{ eV} \] - **(c)** \( n=1 \) to \( n=5 \): \[ \Delta E_{1-5} = E_5 - E_1 = -\frac{13.6}{5^2} - -\frac{13.6}{1^2} = 13.06 \text{ eV} \] - **(d)** \( n=3 \) to \( n=5 \): \[ \Delta E_{3-5} = E_5 - E_3 = -\frac{13.6}{5^2} - -\frac{13.6}{3^2} = 1.55 \text{ eV} \]

Step 5: Compare Energy Differences

To determine which transition requires a longer wavelength than \( 102.6 \text{ nm} \), find which has an energy difference smaller than \( 12.09 \text{ eV} \) (as longer wavelength means less energy). Transitions \( n=2 \) to \( n=4 \) and \( n=3 \) to \( n=5 \) have smaller energy differences (\( 0.85 \text{ eV} \) and \( 1.55 \text{ eV} \) respectively) than \( n=1 \) to \( n=3 \).

Key Concepts

Wavelength and Energy Relationship

Understanding the connection between wavelength and energy is essential when studying transitions in a hydrogen atom. In physics, energy (E) and wavelength (\lambda) are inversely related. This means that as the energy of a photon increases, its wavelength decreases, and vice versa. This relationship is mathematically described by the equation:
\[E = \frac{hc}{\lambda}\]where:

• \(E\) is the energy of the photon (in electron volts, eV).
• \(h\) is Planck's constant (6.626 \times 10^{-34} \text{ Js}).
• \(c\) is the speed of light in a vacuum (3.00 \times 10^8 \text{ m/s}).
• \(\lambda\) is the wavelength of the light (in meters).

This equation indicates that higher energy photons, which are required for larger energy transitions in the hydrogen atom, will have shorter wavelengths. Conversely, transitions with smaller energy differences correspond to photons with longer wavelengths. This principle helps to determine what kind of radiation is required for specific quantum transitions.

Quantum Transitions

Quantum transitions in a hydrogen atom involve an electron moving between different energy levels. Each energy level is identified by a principal quantum number ( n ), which determines the size and energy of the orbital where the electron is likely to be found.
Electrons in the hydrogen atom can absorb or emit photons, allowing them to transition upward or downward between these quantized energy states. When a photon is absorbed, an electron jumps to a higher energy level; while when a photon is emitted, the electron drops to a lower energy level.
For example, moving an electron from the first energy level ( n=1 ) to the third ( n=3 ) requires energy absorption equaling the difference between these two states. Quantum theory describes this phenomenon with precise calculations, allowing us to predict which transitions would require shorter or longer wavelength radiation based on the energy differences.

Energy Level Calculations

Calculating energy levels in a hydrogen atom involves understanding how much energy is associated with each quantum state. Each state is defined by its principal quantum number (n), and the energy for each level can be calculated using the formula:
\[E_n = -\frac{13.6}{n^2} \text{ eV}\]This formula highlights:

• The energy of an electron at a given level (n) is negative—this reflects that energy must be added to free the electron.
• The lower the value of (n), the closer the electron is to the nucleus, and the more negative (or lower) the energy value becomes.

Energy differences (\Delta E) between two levels can be determined by calculating this value for both n_i (initial energy level) and n_f (final energy level), and taking their difference.
For example, for a transition from n=1 to n=3, we calculate:\[\Delta E_{1-3} = E_3 - E_1 = -\frac{13.6}{3^2} - \left(-\frac{13.6}{1^2} \right) = 12.09 \text{ eV}\]Understanding these calculations allows you to determine how much energy and what wavelength of radiation is necessary for specific electron transitions.