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 \(H\) atom. In what region of the spectrum is this radiation found?

Step-by-Step Solution

Answer

Wavelength: \(1.88 \times 10^{-6} \text{ m}\), Frequency: \(1.596 \times 10^{14} \text{ Hz}\), Infrared region.

Step 1: Understanding the Energy Transition

When an electron transitions between energy levels in a hydrogen atom, it emits energy in the form of light. The energy difference between levels can be calculated using the formula: \( \Delta E = E_3 - E_4 = -13.6 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) \) eV, where \( n_3 = 3 \) and \( n_4 = 4 \).

Step 2: Calculating Energy Difference

Substitute \( n_3 = 3 \) and \( n_4 = 4 \) into the formula:\[ \Delta E = -13.6 \left(\frac{1}{3^2} - \frac{1}{4^2}\right) \]\[ \Delta E = -13.6 \left(\frac{1}{9} - \frac{1}{16}\right) \]\[ \Delta E = -13.6 \left(\frac{16 - 9}{144}\right) \]\[ \Delta E = -13.6 \times \frac{7}{144} \approx -0.6611 \text{ eV} \]

Step 3: Converting Energy to Frequency

The energy of the emitted photon is related to its frequency by the equation: \( E = h u \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \). Convert the energy from eV to joules (1 eV = \( 1.602 \times 10^{-19} \text{ J} \)) and solve for \( u \): \[ 0.6611 \text{ eV} = 0.6611 \times 1.602 \times 10^{-19} \text{ J} = 1.058 \times 10^{-19} \text{ J} \]\[ u = \frac{1.058 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 1.596 \times 10^{14} \text{ Hz} \]

Step 4: Calculating Wavelength

The wavelength \( \lambda \) is related to the frequency \( u \) by the equation: \( c = \lambda u \), where \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). Solve for \( \lambda \):\[ \lambda = \frac{c}{u} = \frac{3 \times 10^8}{1.596 \times 10^{14}} \approx 1.88 \times 10^{-6} \text{ m} \]

Step 5: Determining the Spectrum Region

The wavelength \( 1.88 \times 10^{-6} \text{ m} \) corresponds to infrared light, typically in the range of \( 700 \text{ nm} \) to \( 1 \text{ mm} \). Thus, the radiation is in the infrared region of the electromagnetic spectrum.

Key Concepts

Energy levels

Energy levels in a hydrogen atom are defined by the principal quantum number, denoted as \( n \). Each energy level corresponds to a certain orbit of the electron around the nucleus. Notably, these levels are quantized, meaning electrons can only exist at these specific energy points. When an electron moves from a higher energy level to a lower one, this difference in energies results in the emission of light.

For example, when an electron transitions from the fourth energy level, \( n=4 \), to the third energy level, \( n=3 \), it releases energy. This energy can be described using the formula for energy difference:

\[ \Delta E = E_{\text{initial}} - E_{\text{final}} = -13.6 \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \text{ eV} \]
where \( 13.6 \text{ eV} \) is the ionization energy of a hydrogen atom.

Photon emission

Photon emission occurs when an electron drops from a higher energy level to a lower one, sending out a photon – a packet of electromagnetic radiation in the process. This process can be visually represented when electrons in atoms switch orbits.

To understand the photon's characteristics such as frequency or wavelength, one must recognize that the energy released during this transition becomes the energy of the photon. This connection is expressed through the equation:

\[ E = h u \]
Where \( E \) is the photon energy, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), and \( u \) represents the frequency.

This relationship emphasizes the direct conversion of the electron's energy drop into an emitted photon's properties.

Infrared spectrum

The infrared spectrum is part of the electromagnetic spectrum and consists of longer wavelengths compared to visible light. Infrared (IR) radiation is commonly associated with heat, though it is invisible to the human eye.

Infrared light ranges from approximately \( 700 \text{ nm} \) to \( 1 \text{ mm} \) in wavelength. When an electron in a hydrogen atom emits a photon through a transition, the resulting wavelength might fall in the IR region, as was calculated in the original exercise where the wavelength of \( 1.88 \times 10^{-6} \text{ m} \) or \( 1880 \text{ nm} \) is firmly within the infrared range.

• Used in night-vision devices
• Remote controls utilize IR signals
• Essential in various heat-sensing applications

These emissions highlight the broad spectrum of electromagnetic radiation beyond visible light.

Electron transition calculations

Electron transition calculations are crucial for determining the specific characteristics of the emitted photon. In hydrogen atoms, these calculations explain how electrons move between different energy levels and the corresponding electromagnetic radiation produced.

The process involves using known formulas, such as to calculate the energy difference and then converting this energy into terms of frequency or wavelength.
Steps include:

• Calculate the energy difference \( \Delta E \)
• Convert \( \Delta E \) from electron volts (eV) to joules (J)
• Determine frequency \( u \) using \( E = h u \)
• Find wavelength \( \lambda \) from \( \lambda = \frac{c}{u} \)

These steps allow us to pinpoint exactly where this transition stands within the electromagnetic spectrum, be it infrared, visible, or another region entirely.