Question

One of the visible lines in the hydrogen emission spectrum corresponds to the \(n=6\) to \(n=2\) electronic transition. What color light is this transition? See Exercise \(138 .\)

Short answer

The transition from \(n=6\) to \(n=2\) in the hydrogen emission spectrum corresponds to a wavelength of \(244.57\, \text{nm}\), which falls into the ultraviolet (UV) region of the spectrum. Although it is not technically visible to the human eye, it is very close to the violet end of the visible spectrum (approximately 380-450 nm). Therefore, the emitted light would be a very deep violet or almost ultraviolet color.

Step-by-step solution

Recall the Rydberg formula for hydrogen

The Rydberg formula for hydrogen is given by: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant for hydrogen (\(1.097 \times 10^7 \, \text{m}^{-1}\)), \(n_1\) is the smaller principal quantum number, and \(n_2\) is the larger principal quantum number.

Plug in the quantum numbers for the electronic transition

For the given transition from \(n=6\) to \(n=2\), we have \(n_1 = 2\) and \(n_2 = 6\). Plugging these values into the Rydberg formula, we have: \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \)

Calculate the wavelength of emitted light

Substituting the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)) into the equation and solving for \( \lambda \), we get: \( \frac{1}{\lambda} = (1.097 \times 10^7 \, \text{m}^{-1}) \left( \frac{1}{4} - \frac{1}{36} \right) \) \( \frac{1}{\lambda} = 4.087 \times 10^6 \, \text{m}^{-1} \) So, the wavelength (\( \lambda \)) of emitted light is: \( \lambda = \frac{1}{4.087 \times 10^6 \, \text{m}^{-1}} \) \( \lambda = 244.57\, \text{nm} \)

Determine the color of light based on the wavelength

With the given wavelength of \(244.57\, \text{nm}\), we need to identify which color in the visible spectrum it corresponds to. The wavelength falls into the ultraviolet (UV) region of the spectrum, which is technically not visible to the human eye. However, it is very close to the violet end of the visible spectrum (approximately 380-450 nm), so if we were to approximate, we could say that the emitted light would be a very deep violet or almost ultraviolet color.

Key concepts

Rydberg Formula

The Rydberg formula is an essential piece of the puzzle when it comes to understanding the emission spectrum of hydrogen. It's an equation used to predict the wavelength of light resulting from an electron moving between energy levels in a hydrogen atom.

This formula is written as \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where:\

• \(\lambda\) represents the wavelength of the emitted light,
• \(R_H\) is the Rydberg constant specific to hydrogen, usually expressed as \(1.097 \times 10^7 \, \text{m}^{-1}\),
• \(n_1\) is the lower energy level, and
• \(n_2\) is the higher energy level the electron transitions from.

Using the Rydberg formula, we can calculate the specific wavelengths for the spectral lines of hydrogen, which are part of its unique fingerprint.

Electronic Transition

When discussing the hydrogen emission spectrum, it's crucial to grasp what an 'electronic transition' is. Essentially, this term refers to an electron's jump between different energy levels or orbits within an atom.

When an electron absorbs energy, it can jump up to a higher energy level, and when it falls back down to a lower energy level, it emits energy in the form of light. The energy difference between these levels determines the wavelength of light emitted, and this is accurately predicted by the Rydberg formula.

For example, an electron transitioning from the \(n=6\) to the \(n=2\) energy level in a hydrogen atom will release a photon with a specific wavelength that we can calculate.

Wavelength Calculation

To find the wavelength of light emitted during an electronic transition, we use the equation derived from the Rydberg formula. As demonstrated in the problem, we input the initial and final energy levels, along with the Rydberg constant, into the formula.

Here's the stepwise calculation process:

Insert Quantum Numbers

First, we insert the principal quantum numbers corresponding to the energy levels involved in the transition into the formula.

Apply the Rydberg Constant

Next, we input the Rydberg constant for hydrogen.

Perform the Calculation

After substituting these values, we rearrange the formula to solve for \(\lambda\), which represents the wavelength of the light. Solving this gives us the exact numerical value of the wavelength corresponding to the energy transition in question.

Visible Spectrum

The visible spectrum is the portion of the electromagnetic spectrum that the human eye can perceive. It ranges approximately from 380 nm to 750 nm and includes all the colors that we can see – violet, blue, green, yellow, orange, and red.

Although the exact boundaries of the visible spectrum can vary from person to person, any light with a wavelength shorter than this range falls into the ultraviolet (UV) region, while longer wavelengths are part of the infrared (IR) region.

In our case, a wavelength of \(244.57\, \text{nm}\) lies just beyond the violet end of the visible spectrum. Therefore, we categorize it as ultraviolet, and it's not visible to the naked eye. If it were within the visible spectrum, we would describe the color as a deep violet due to its proximity to the visible range.