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 \(103 \, \text{nm}\), frequency is \(2.92 \times 10^{15} \, \text{Hz}\), and this lies in the UV spectrum.

Step 1: Identify the Transition Energy

The energy difference between two levels in a hydrogen atom is given by the formula \( E = -13.6 \times \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \) eV, where \( n_1\) and \( n_2 \) are the principal quantum numbers of the two levels. Here, \( n_2 = 3 \) and \( n_1 = 1 \). Calculate \( E \):\[ E = -13.6 \times \left(\frac{1}{1^2} - \frac{1}{3^2}\right) \]\[ E = -13.6 \times \left(1 - \frac{1}{9}\right) = -13.6 \times \frac{8}{9} = -12.1 \text{ eV} \] (transition energy in electron volts).

Step 2: Convert Energy to Joules

To find the frequency and wavelength, first convert the energy from electron volts to joules by multiplying with the conversion factor, \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). \[ E = -12.1 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = -1.938 \times 10^{-18} \, \text{J} \]

Step 3: Calculate the Frequency

Use the energy-frequency relation \( E = h \cdot f \), where \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck's constant, to determine the frequency \( f \).\[ f = \frac{E}{h} = \frac{1.938 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \approx 2.92 \times 10^{15} \, \text{Hz} \]

Step 4: Calculate the Wavelength

Use the speed of light equation \( c = \lambda \cdot f \) (where \( c = 3.00 \times 10^8 \, \text{m/s} \)) to find the wavelength \( \lambda \).\[ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \, \text{m/s}}{2.92 \times 10^{15} \, \text{Hz}} \approx 1.03 \times 10^{-7} \, \text{m} = 103 \, \text{nm} \]

Step 5: Determine Spectrum Region

The calculated wavelength \( 103 \, \text{nm} \) is within the ultraviolet (UV) region of the electromagnetic spectrum, as UV range is approximately from 10 to 400 nm.

Key Concepts

Electron Transition

In the realm of hydrogen atom spectroscopy, electron transitions are pivotal events. When an electron in a hydrogen atom moves between different energy levels, it absorbs or emits energy in the form of light. Each energy level in an atom is associated with a quantum number, denoted as \(n\). In the given problem, the electron transitions from a higher energy level, \(n=3\), to a lower energy level, \(n=1\). This transition results in the release of energy since the electron is moving to a more stable, lower energy state.

• Higher quantum number (e.g., \(n=3\)): Higher energy, less stable.
• Lower quantum number (e.g., \(n=1\)): Lower energy, more stable.
• Energy is emitted when the electron drops to a lower energy level.

Understanding these transitions is essential as they form the basis of spectral lines observed in spectroscopy.

Energy Levels

Energy levels in hydrogen atom are quantized, meaning an electron can only exist at specific energy states, not in between. These levels are defined by the principal quantum number \(n\), which is a positive integer. The energy for each level is calculated using the formula: \[ E_n = -13.6 imes \frac{1}{n^2} \text{ eV} \] where \(-13.6\, \text{eV}\) is the ionization energy of hydrogen. As \(n\) increases, the energy levels get closer together. For an electron transitioning from \(n=3\) to \(n=1\), the energy difference is calculated using: \[ E = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV} \]This difference (or energy release) determines the frequency of the light emitted. Understanding energy levels helps in predicting the possible transitions and the corresponding spectral lines in hydrogen atom.

Wavelength Calculation

To calculate the wavelength of light emitted during an electron transition, we utilize energy-frequency-wavelength relationships. Once the energy for the transition is known, it can be converted into wavelength through a series of steps involving basic equations.

First, convert the energy from electron volts to joules. Then, use the relationship between energy \(E\) and frequency \(f\):\[ f = \frac{E}{h} \]Here, \(h\) is Planck’s constant \((6.626 \times 10^{-34} \text{ J}\cdot\text{s})\). Once frequency is known, calculate the wavelength \(\lambda\) using the speed of light \(c\):\[ \lambda = \frac{c}{f} \]where \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\).

• Calculate frequency from energy.
• Calculate wavelength from frequency.
• Conclusions about spectrum regions from wavelength.

Providing a concrete understanding of how energy translates into light properties.

Ultraviolet Spectrum

The ultraviolet (UV) spectrum is a part of the electromagnetic spectrum with wavelengths between approximately 10 nm and 400 nm. In our problem, the calculated wavelength of the emitted light is 103 nm, which clearly places it in the UV range. This region is beyond the visible spectrum and is characterized by higher energy photons.

• UV radiation is less visible to the human eye.
• Occurs at shorter wavelengths with higher energy than visible light.
• Plays a role in causing chemical reactions, such as those needed for vitamin D synthesis in skin.

Understanding where a wavelength falls on the spectrum helps in identifying the type of radiation and its possible applications or effects, such as germicidal properties and its importance in astronomy for studying celestial objects. The UV classification in our solution shows the energy level changes in the hydrogen atom are significant, affecting electronic transitions and resulting in specific spectral emissions.