Problem 69 - Chemistry and Chemical Reactivity

Question

For an electron in a hydrogen atom, calculate the energy of the photon emitted when an electron falls in energy from the \(n=5\) level to the \(n=2\) state. What are the frequency and wavelength of this electromagnetic radiation?

Step-by-Step Solution

Answer

Energy is \(4.57 \times 10^{-19} \, \text{J}\), frequency is \(6.89 \times 10^{14} \, \text{Hz}\), and wavelength is \(435 \, \text{nm}\).

Step 1: Identify Energy Level Transition

We need to calculate the energy emitted by an electron transitioning from the \(n=5\) level to the \(n=2\) level in a hydrogen atom. The formula to find the energy of emitted photon during a transition between two levels is given by the Rydberg Formula: \( E = -13.6 \, \text{eV} \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \), where \( n_{\text{final}} = 2\) and \( n_{\text{initial}} = 5\).

Step 2: Calculate the Energy of the Photon

Plug the values of \(n_{\text{initial}}\) and \(n_{\text{final}}\) into the Rydberg Formula:\[E = -13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = -13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{25} \right)\]Simplify the expression within parentheses:\[\frac{1}{4} - \frac{1}{25} = \frac{25}{100} - \frac{4}{100} = \frac{21}{100}\]So, the energy is:\[E = -13.6 \, \text{eV} \times \frac{21}{100} = -2.856 \, \text{eV}\]The negative sign indicates that the energy is released.

Step 3: Convert Energy to Joules

We need to convert the energy from electron volts to joules since it's standard to work with joules in SI units. Use the conversion: \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\).\[E = -2.856 \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = -4.57 \times 10^{-19} \, \text{J}\]

Step 4: Calculate Frequency of the Photon

Using the energy of the photon calculated, we apply the formula \(E = h u\) where \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\) and \(u\) is frequency.\[u = \frac{E}{h} = \frac{4.57 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} = 6.89 \times 10^{14} \, \text{Hz}\]

Step 5: Calculate Wavelength of the Photon

To find the wavelength \( \lambda \), use the speed of light \( c = 3.00 \times 10^8 \, \text{m/s}\) combined with the frequency \( u\) from the previous step, in the equation \(c = \lambda u\).\[\lambda = \frac{c}{u} = \frac{3.00 \times 10^8 \, \text{m/s}}{6.89 \times 10^{14} \, \text{Hz}} = 4.35 \times 10^{-7} \, \text{m} = 435 \, \text{nm}\]

Step 6: Interpret the Results

The energy of the photon emitted in this transition is \(-2.856 \, \text{eV}\), which implies a photon of approximately \(4.57 \times 10^{-19} \, \text{J}\). The frequency of this photon is \(6.89 \times 10^{14} \, \text{Hz}\), and its wavelength is about \(435 \, \text{nm}\), which is within the visible spectrum range.

Key Concepts

Rydberg Formula

The Rydberg formula is central to understanding atomic transitions in hydrogen. It allows us to calculate the energy difference between two electron energy levels. This formula is particularly important in quantum mechanics and atomic physics. Rydberg's formula is represented as:

• \( E = -13.6 \, \text{eV} \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \)

This equation helps us determine the energy released or absorbed when an electron moves between energy levels:

• \(n_{\text{initial}}\) is the energy level where the electron starts.
• \(n_{\text{final}}\) is the energy level where the electron ends up.

For example, if an electron falls from the 5th energy level to the 2nd, the formula calculates the exact energy change, providing insight into the properties of emitted or absorbed electromagnetic radiation. It's useful for visualizing quantum jumps within atoms.
By plugging in the values, we can ascertain that a transition from \(n=5\) to \(n=2\) results in a specific energy change that ultimately determines the characteristics of the light wave emitted.

Photon Energy Calculation

Calculating the energy of a photon emitted during an electron transition relies on the basic principles of energy conservation and quantum physics. Using the Rydberg formula, we computed that the energy released during a transition from the 5th to the 2nd energy level is \(-2.856 \, \text{eV}\).
This negative sign signifies energy release, which is typical when electrons drop to lower energy levels. To convert this energy into joules, a standard SI unit of energy, we use the conversion factor:

• \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\)

Thus, the calculation becomes:

• \(-2.856 \, \text{eV} \times 1.602 \times 10^{-19} = -4.57 \times 10^{-19} \, \text{J}\)

This photon's energy is pivotal in defining its frequency and wavelength, leading to further calculations of its physical properties.

Electromagnetic Radiation

Electromagnetic radiation occurs when an energetic process, like an electron transition, releases energy in the form of a photon. This process is key in understanding phenomena such as light emission and absorption in atoms.
Light, or electromagnetic radiation, is characterized by its dual wave-particle nature. It can be described by:

• Frequency (\(u\))
• Wavelength (\(\lambda\))

The relationship between frequency and wavelength is governed by the speed of light equation:

• \(c = \lambda u\)

Here, \(c\) is the speed of light, approximately \(3.00 \times 10^8\, \text{m/s}\).
Calculating these values allows us to classify the type of radiation emitted. In our case study, the wavelength falls within the visible spectrum, showing how transitions in atoms are visible to us as light colors. It's these kinds of transitions that make phenomena like fireworks and stellar spectra so colorful and intriguing.

Energy Level Transitions

Energy level transitions in atoms are direct manifestations of quantum mechanics at work. In a hydrogen atom, electrons occupy distinct energy levels. These levels are analogous to steps in a staircase that an electron can hop between, but only with the exact right amount of energy.

• When an electron moves to a lower energy level, it emits energy equivalent to the difference between the two levels.
• This emitted energy is released in the form of a photon, a particle of light.

These quantum leaps between levels are responsible for many observable properties in the universe, from the colors of neon lights to the absorption lines seen in the spectra of distant stars.
What's fascinating is that each element displays its unique set of energy levels, leading to characteristic spectrums termed spectral lines. By examining these transitions, scientists can deduce a lot about the environment and composition of substances, making this concept a cornerstone of spectroscopy.