Problem 21 - Chemistry and Chemical Reactivity

Question

Calculate the wavelength and frequency of light emitted when an electron changes from \(n=3\) to \(n=1\) in the \(\mathrm{H}\) atom. In what region of the spectrum is this radiation found?

Step-by-Step Solution

Answer

The wavelength is 103 nm (ultraviolet) with a frequency of \(2.91 \times 10^{15}\) Hz.

Step 1: Understand the Energy Transition Formula

For hydrogen atoms, the energy difference between levels can be calculated using the formula \( \Delta E = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \), where \( R_H = 2.18 \times 10^{-18} \text{ J} \) is the Rydberg constant, \( n_i \) is the initial energy level, and \( n_f \) is the final energy level.

Step 2: Calculate the Energy Difference

Plug in the given values into the energy transition formula: \[ \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{3^2} - \frac{1}{1^2} \right) = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{9} - 1 \right) \]. Calculate the result to find \( \Delta E \).

Step 3: Compute the Energy Change

Solve the expression: \[ \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{9} - 1 \right) \], which simplifies to \[ \Delta E = 2.18 \times 10^{-18} \text{ J} \times \left( -\frac{8}{9} \right) = -1.94 \times 10^{-18} \text{ J} \]. The negative sign indicates the emission of a photon.

Step 4: Use the Energy-Wavelength Relation

Relate energy to wavelength using \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant and \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. Rearrange to find \( \lambda = \frac{hc}{|\Delta E|} \).

Step 5: Calculate the Wavelength

Substitute the known values into the wavelength formula: \[ \lambda = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m/s}}{1.94 \times 10^{-18} \text{ J}} \]. Evaluate to find \( \lambda \).

Step 6: Compute the Wavelength Result

Perform the calculation: \[ \lambda \approx 1.03 \times 10^{-7} \text{ m} \] or \( 103 \text{ nm} \), which is in the ultraviolet region.

Step 7: Wavelength to Frequency Conversion

Use the relation \( c = \lambda u \) to find frequency \( u \), rearranging gives \( u = \frac{c}{\lambda} \).

Step 8: Calculate the Frequency

Insert the values into the frequency formula: \[ u = \frac{3.00 \times 10^8 \text{ m/s}}{1.03 \times 10^{-7} \text{ m}} \]. Compute to obtain the frequency \( u \).

Step 9: Compute the Frequency Result

Calculate \( u \approx 2.91 \times 10^{15} \text{ Hz} \).

Step 10: Identify the Spectrum Region

The wavelength \( 103 \text{ nm} \) indicates that the radiation is in the ultraviolet region of the electromagnetic spectrum.

Key Concepts

Wavelength Calculation

When an electron transitions between energy levels in a hydrogen atom, it either absorbs or emits light. To calculate the wavelength of the emitted light, we use the relation between energy and wavelength. This is expressed by the formula:
\[\lambda = \frac{hc}{|\Delta E|}\]where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ Js} \), \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( |\Delta E| \) is the absolute value of the energy change calculated in the transition.
For an electron transitioning from \( n=3 \) to \( n=1 \) in a hydrogen atom, substitute the known constants into the equation to find the wavelength \( \lambda \). This results in a wavelength of approximately \( 103 \text{ nm} \), which places the radiation in the ultraviolet region of the electromagnetic spectrum.

• The method involves determining energy change first.
• The formula rearranges to solve directly for wavelength.
• Wavelength provides clues to the type of light emitted.

Frequency Calculation

Frequency and wavelength are inversely related. Once the wavelength of emitted light is found, we can calculate its frequency using the speed of light formula:
\[u = \frac{c}{\lambda}\]where \( u \) is frequency, \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength.
Using the wavelength calculated earlier \( 103 \text{ nm} \), the frequency is determined by plugging these values into the equation. The result is approximately \( 2.91 \times 10^{15} \text{ Hz} \).

• Frequency informs about the type of wave.
• High frequency corresponds to high energy radiation.
• Understanding these relationships helps categorize light.

Rydberg Constant

The Rydberg constant is crucial in calculations involving hydrogen's spectral lines. It is denoted as \( R_H \) and is approximately \( 2.18 \times 10^{-18} \text{ J} \). This constant represents the limits of spectroscopic analysis for hydrogen.
To find the energy difference during an electron transition, we use:
\[\Delta E = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)\]where \( n_i \) and \( n_f \) are the initial and final energy levels. For a transition from \( n=3 \) to \( n=1 \), plugging in these values yields the energy change. This change is then used to find wavelength and frequency.

• Essential for predicting spectral lines.
• Applies to electron transitions in hydrogen.
• Connects quantum mechanics to observable phenomena.

Ultraviolet Radiation

Ultraviolet (UV) radiation is part of the electromagnetic spectrum with wavelengths shorter than visible light but longer than X-rays. The calculated wavelength of \( 103 \text{ nm} \) for an electron moving from \( n=3 \) to \( n=1 \) in hydrogen places it squarely in the UV range.
UV radiation has several key characteristics:

• It is responsible for causing fluorescence in certain materials.
• UV light has higher energy than visible light, making it capable of causing sunburns.
• It plays a role in chemical reactions, such as vitamin D synthesis in humans.

Understanding the nature of UV radiation helps explain why certain electron transitions result in light not visible to the naked eye but significant in scientific observations.