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 the level with \(n=1\) to the level with \(n=3\) 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 \(n=4\) (c) \(n=1\) to \(n=5\) (b) \(n=1\) to \(n=4\) (d) \(n=3\) to \(n=5\)

Step-by-Step Solution

Answer

Transition from \(n=3\) to \(n=5\) requires longer wavelength radiation.

Step 1: Understand the Energy-Wavelength Relationship

The energy difference between two levels in a hydrogen atom is related to the wavelength of radiation required by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy, \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. A longer wavelength corresponds to lower energy.

Step 2: Recall the Formula for Energy Levels in Hydrogen

The energy levels of a hydrogen atom are given by the formula \( E_n = -\frac{13.6}{n^2} \) eV, where \( n \) is the principal quantum number. The energy of the photon needed for a transition between two levels \( n_1 \) and \( n_2 \) is \( \Delta E = E_{n_2} - E_{n_1} \).

Step 3: Calculate the Energy Required for Transitions

For each option, calculate the energy difference \( \Delta E \) between the initial and final states using the formula: - For (a) from \( n=2 \) to \( n=4 \), \( \Delta E = -13.6 \left( \frac{1}{4^2} - \frac{1}{2^2} \right) \) - For (b) from \( n=1 \) to \( n=4 \), \( \Delta E = -13.6 \left( \frac{1}{4^2} - \frac{1}{1^2} \right) \) - For (c) from \( n=1 \) to \( n=5 \), \( \Delta E = -13.6 \left( \frac{1}{5^2} - \frac{1}{1^2} \right) \) - For (d) from \( n=3 \) to \( n=5 \), \( \Delta E = -13.6 \left( \frac{1}{5^2} - \frac{1}{3^2} \right) \)

Step 4: Compare Energy Differences to Determine Wavelengths

Calculate each \( \Delta E \) and relate it to the wavelength using \( \lambda = \frac{hc}{\Delta E} \). The option with the smallest \( \Delta E \) will have the longest wavelength.- For (a), after calculation, \( \Delta E \approx 2.55 \) eV- For (b), \( \Delta E \approx 12.75 \) eV- For (c), \( \Delta E \approx 13.14 \) eV- For (d), \( \Delta E \approx 0.967 \) eV

Step 5: Conclusion – Longest Wavelength Transition

The transition from \( n=3 \) to \( n=5 \) (option d) has the smallest energy difference and thus requires radiation of the longest wavelength. This meets the condition of requiring radiation with a longer wavelength than the transition from \( n=1 \) to \( n=3 \).

Key Concepts

Energy-Wavelength Relationship

When discussing hydrogen atom transitions, the relationship between energy and wavelength is crucial. To put it simply, energy and wavelength have an inverse relationship, meaning when one goes up, the other comes down. This relationship is captured by the equation:\[E = \frac{hc}{\lambda}\]where:

• \(E\) is the energy of the photon,
• \(h\) is Planck's constant, a fundamental value in quantum mechanics,
• \(c\) is the speed of light,
• \(\lambda\) is the wavelength of light.

From this equation, you can see how energy changes with wavelength. A longer wavelength corresponds to lower energy, and a shorter wavelength means higher energy.
This inverse relationship allows us to determine what type of radiation is associated with different energy transitions. For example, visible light has a certain range of wavelengths, and its color depends on these wavelengths. Longer wavelengths in the visible spectrum correspond to red light, while shorter wavelengths correspond to violet light.

Principal Quantum Number

The principal quantum number, often represented by \(n\), is fundamental to understanding hydrogen atom energy levels. It determines the energy and size of an electron's orbit around the nucleus.
Electrons reside in energy levels that are quantized, meaning they can only have specific energy values, determined by this principal quantum number. The energy levels are further apart as the principal quantum number increases.
In hydrogen, the formula for energy levels is given by:\[E_n = -\frac{13.6}{n^2} \text{ eV}\]This formula shows that the energy is:

• Inversely proportional to the square of the principal quantum number.
• Negative, indicating that the electron is bound to the atom.

For the hydrogen atom:

• The ground state, or lowest energy level, is \(n=1\).
• Higher energy levels are \(n=2, 3, 4, ...\) and so on.

Therefore, as the principal quantum number increases, the electron is in a higher energy state with more energy.

Photon Energy Calculation

Calculating the energy of a photon involved in a transition between two energy levels provides insights into the kind of radiation involved. The energy difference, \(\Delta E\), between two levels is calculated as follows:\[\Delta E = E_{n_2} - E_{n_1}\]The photon must have this exact energy to cause a transition from \(E_{n_1}\) to \(E_{n_2}\).
Let's break it down:

• Identify the initial and final states using their principal quantum numbers.
• Substitute these values into the energy formula for each state's energy.
• Compute \(\Delta E\), which tells us how much energy the photon should have.

For example, transitioning from \(n=2\) to \(n=4\) is calculated using:\[\Delta E = -13.6 \left( \frac{1}{4^2} - \frac{1}{2^2} \right) \text{ eV}\]Perform the subtraction inside the bracket first, convert it with the constant, and you will find the amount of energy required for the transition. If you need the wavelength, simply rearrange the energy-wavelength equation, solving for \(\lambda\):\[\lambda = \frac{hc}{\Delta E}\]This approach ties together all the core concepts: energy differences, quantum numbers, wavelengths, and photon energies.