Problem 3 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

What is the wavelength of light for a line in the atomic spectrum of hydrogen for which \(n_{1}=2\) and \(n_{2}=4 ?\) What part of the electromagnetic spectrum does this correspond to? (Section 3.3)

Step-by-Step Solution

Answer

The wavelength is 486 nm, which corresponds to blue light in the visible spectrum.

Step 1: Understand the Rydberg Formula

The Rydberg formula is used to calculate the wavelengths of spectral lines in many chemical elements. For hydrogen, it is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \) is the Rydberg constant for hydrogen, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels.

Step 2: Substitute Known Values

Substitute the known values into the Rydberg formula: \( n_1 = 2 \) and \( n_2 = 4 \). Thus, \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] Simplify the expression: \( \frac{1}{2^2} = \frac{1}{4} \) and \( \frac{1}{4^2} = \frac{1}{16} \).

Step 3: Calculate the Difference in Fractions

Calculate \( \frac{1}{4} - \frac{1}{16} \): \[ \frac{1}{4} = \frac{4}{16} \] \[ \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]

Step 4: Calculate the Wavelength Inverse

Substitute the fraction back into the equation: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16} \] Simplify the expression: \[ \frac{1}{\lambda} = 2.056875 \times 10^6 \text{ m}^{-1} \]

Step 5: Solve for Wavelength

Find \( \lambda \) by taking the reciprocal: \[ \lambda = \frac{1}{2.056875 \times 10^6} \] Simplifying, \( \lambda \approx 4.86 \times 10^{-7} \text{ m} \) or \( 486 \text{ nm} \). This wavelength corresponds to the visible spectrum, specifically in the blue part.

Key Concepts

Rydberg formula

The Rydberg formula is a crucial tool used in physics and chemistry for determining the wavelengths of light emitted or absorbed by electrons in atoms. For hydrogen, this formula is particularly important because it helps explain the unique spectral lines observed in its atomic spectrum.
The formula is expressed as:

• \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Here, \( \lambda \) represents the wavelength of the light emitted or absorbed, \( R_H \) is the Rydberg constant for hydrogen, and \( n_1 \) and \( n_2 \) are the principal quantum numbers representing the electron's initial and final energy levels. The Rydberg constant, \( R_H \), has a value of approximately \( 1.097 \times 10^7 \text{ m}^{-1} \). This formula accounts for the energy changes that occur when an electron transitions between energy levels, leading to specific wavelengths of light being observed in the spectrum.

Principal quantum numbers

Principal quantum numbers, often denoted as \( n \), are fundamental in describing an electron's position in an atom. They play a crucial role in determining the energy levels of electrons.

• Each principal quantum number, \( n \), corresponds to a specific electron shell or energy level.
• The values of \( n \) can be any positive integer \( n = 1, 2, 3, ... \).
• A higher \( n \) value means the electron is further from the nucleus and possesses higher energy.

In the Rydberg formula, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states of the electron, respectively. As an electron absorbs or emits energy, it transitions between these states, resulting in spectral lines that can be measured using the Rydberg formula.

Wavelength calculation

Calculating the wavelength of light emitted or absorbed in an atomic transition is a straightforward process using the Rydberg formula. Let's break it down:

• First, substitute the known principal quantum numbers \( n_1 \) and \( n_2 \) into the Rydberg formula.
• Calculate the differences in the squared reciprocals: \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \).
• Apply the Rydberg constant \( R_H \) to find \( \frac{1}{\lambda} \).
• Take the reciprocal to solve for \( \lambda \), which is the wavelength.

For example, substituting \( n_1 = 2 \) and \( n_2 = 4 \) into the formula gives \[ \lambda \approx 486 \text{ nm} \]. This result positions the wavelength within the visible spectrum, often appearing as blue light.

Electromagnetic spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, from very short gamma rays to very long radio waves. The human eye can only perceive a small range of this spectrum, known as visible light.
Wavelengths in the electromagnetic spectrum are typically measured in meters, nanometers, or Angstroms, spanning a broad range of wavelengths and frequencies.

• Gamma Rays: Shortest wavelengths, typically less than 0.01 nanometers.
• X-Rays: Wavelengths ranging from 0.01 to 10 nanometers.
• Ultraviolet: Extends from about 10 to 400 nanometers.
• Visible Light: Ranges from 400 (violet) to 700 nanometers (red).
• Infrared: Wavelengths from 700 nanometers to 1 millimeter.
• Microwave: Spans from 1 millimeter to 1 meter.
• Radio Waves: Largest wavelengths, ranging from 1 meter and beyond.

When calculating the wavelength of a hydrogen spectral line, resulting data often falls within the visible range. In the solved example, 486 nm signifies blue light, exemplifying how atomic transitions contribute to our understanding of color within the electromagnetic spectrum.