Problem 22 - Chemistry and Chemical Reactivity

Question

Calculate the wavelength and frequency of light emitted when an electron changes from \(n=4\) to \(n=3\) in the \(\mathrm{H}\) atom. In what region of the spectrum is this radiation found?

Step-by-Step Solution

Answer

The emitted light is in the infrared region with a wavelength of about 1.87 µm and frequency of 1.60 x 10¹⁴ Hz.

Step 1: Determine the energy change

The energy difference between two levels in the hydrogen atom can be calculated using the formula: \[ \Delta E = -13.6 \text{ eV} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]where \( n_1 = 4 \) and \( n_2 = 3 \). Substitute the values into the equation:\[ \Delta E = -13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \text{ eV} = -13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \text{ eV} \]\[ \Delta E = -13.6 \left( \frac{16 - 9}{144} \right) \text{ eV} = -13.6 \left( \frac{7}{144} \right) \text{ eV} \]\[ \Delta E \approx -0.661 \text{ eV} \]

Step 2: Convert energy to wavelength

Use the relation between energy and wavelength:\[ E = \frac{hc}{\lambda} \]Where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3.00 \times 10^8 \text{ m/s} \). First, convert energy from eV to Joules:\[ 0.661 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} \approx 1.06 \times 10^{-19} \text{ J} \]Then, solve for \( \lambda \):\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.06 \times 10^{-19}} \]\[ \lambda \approx 1.87 \times 10^{-6} \text{ m} \]

Step 3: Calculate the frequency

Use the formula for speed of light \( c = \lambda u \), where \( u \) is the frequency. Solve for \( u \):\[ u = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{1.87 \times 10^{-6}} \]\[ u \approx 1.60 \times 10^{14} \text{ Hz} \]

Step 4: Determine the spectrum region

Compare the calculated wavelength with known regions of the electromagnetic spectrum. A wavelength of \( 1.87 \times 10^{-6} \text{ m} \) corresponds to the infrared region of the spectrum.

Key Concepts

Energy Levels

In a hydrogen atom, electrons naturally exist in specific energy levels, denoted by the principal quantum number \( n \). These energy levels can be imagined like steps on a ladder where each step represents a different energy state. As we climb higher up the steps, or energy levels, the amount of energy associated with each level increases. When an electron transitions between these levels, it either absorbs or emits a specific amount of energy.
The formula for the energy difference between two levels \( n_1 \) and \( n_2 \) in a hydrogen atom is:

• \( \Delta E = -13.6 \text{ eV} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \)

This formula helps us calculate the energy required for an electron to jump between two specified energy levels. In our example, an electron transitioning from \( n=4 \) to \( n=3 \), results in the release of approximately \(-0.661\text{ eV}\) as it moves to a lower energy state.
Understanding these transitions is crucial because they lay the foundation for the other calculations related to wavelength and frequency.

Wavelength Calculation

Once we have the energy change, \( \Delta E \), we can move on to calculating the wavelength of the emitted light. The energy of a photon is directly related to its wavelength through the equation:

• \( E = \frac{hc}{\lambda} \)

Here, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \) and \( c \) is the speed of light \( (3.00 \times 10^8 \text{ m/s}) \). To find the wavelength \( \lambda \), we simply rearrange the formula:

• \( \lambda = \frac{hc}{E} \)

For our electron transition from \( n=4 \) to \( n=3 \), we calculated the energy in Joules ( \( \approx 1.06 \times 10^{-19} \text{ J} \)) and substituted it into the formula to find the wavelength. The resulting wavelength was approximately \( 1.87 \times 10^{-6} \text{ m} \).
By knowing the wavelength, we can understand more about the type of light emitted, as we'll see in the section about the electromagnetic spectrum.

Frequency Calculation

The next step involves calculating the frequency of the emitted light. Frequency and wavelength are related by the speed of light, according to the formula:

• \( c = \lambda \cdot u \)

Where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( u \) is the frequency. After finding the wavelength, we rearrange the standard equation to solve for the frequency:

• \( u = \frac{c}{\lambda} \)

Substituting our calculated wavelength \( 1.87 \times 10^{-6} \text{ m} \) into this formula gives us a frequency of about \( 1.60 \times 10^{14} \text{ Hz} \).
Frequency is a crucial aspect of understanding electromagnetic radiation, as it helps determine the light's characteristics and interactions, as we will explore further in terms of spectrum analysis.

Electromagnetic Spectrum

The electromagnetic spectrum is the range of all types of electromagnetic radiation. Light emitted or absorbed during electron transitions falls somewhere on this spectrum, which includes forms such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each category is defined by its specific range of wavelengths and frequencies.
In our electron transition example, the calculated wavelength \( 1.87 \times 10^{-6} \text{ m} \) places the emitted light within the infrared region of the spectrum. The infrared range generally covers wavelengths from about \( 700 \text{ nm} \) to \( 1 \text{ mm} \).

• Wavelength below \( 700 \text{ nm} \) enters the visible spectrum
• Wavelength above \( 1 \text{ mm} \) enters the microwave spectrum

This classification allows scientists to determine the kind of radiation they are dealing with and understand its possible applications in fields like communication, astronomy, and medicine. Knowing the region of the electromagnetic spectrum is important for applying the appropriate technology or investigation techniques.