Question

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

Short answer

The wavelengths of light emitted during the given electron transitions in the hydrogen atom are as follows: a. Transition from \(n=3\) to \(n=2\): \(\lambda = 656.46 \, nm\) (visible light, red) b. Transition from \(n=4\) to \(n=2\): \(\lambda = 486.24 \, nm\) (visible light, blue-green) c. Transition from \(n=2\) to \(n=1\): \(\lambda = 121.60 \, nm\) (ultraviolet light)

Step-by-step solution

Recall the Rydberg formula for hydrogen atoms

The Rydberg formula gives the wavelength of light emitted or absorbed during electron transitions in a hydrogen atom. It is represented as: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where: - \(\lambda\) is the wavelength of the emitted light - \(R_H\) is the Rydberg constant for hydrogen (\(R_H = 1.097 \times 10^7 \, m^{-1}\)) - \(n_1\) is the initial energy level of the electron - \(n_2\) is the final energy level of the electron We'll use this formula to calculate the wavelength of light emitted during each transition.

Calculate the wavelength for each transition

To calculate the wavelength of each transition, we'll substitute the initial and final energy levels into the Rydberg formula and solve for \(\lambda\). a. Transition from \(n=3\) to \(n=2\): \(\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\) \(\lambda = 656.46 \, nm\) b. Transition from \(n=4\) to \(n=2\): \(\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{2^2} - \frac{1}{4^2}\right)\) \(\lambda = 486.24 \, nm\) c. Transition from \(n=2\) to \(n=1\): \(\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\) \(\lambda = 121.60 \, nm\)

Identify the type of electromagnetic radiation

Now that we have the wavelength for each transition, we can determine the type of electromagnetic radiation using the electromagnetic spectrum: 1. \(656.46 \, nm\) falls into the visible light range, specifically in the red color range. 2. \(486.24 \, nm\) also falls into the visible light range, specifically in the blue-green color range. 3. \(121.60 \, nm\) falls into the ultraviolet (UV) range. So, the types of electromagnetic radiation emitted during each transition are: a. Visible light (red) b. Visible light (blue-green) c. Ultraviolet light

Key concepts

Electromagnetic Radiation

Electromagnetic radiation (EMR) covers a wide range of wavelengths and frequencies. This spectrum includes various types like radio waves, microwaves, infrared, visible light, ultraviolet (UV) rays, X-rays, and gamma rays.

- **Wavelength** equates to the distance between identical points in two consecutive waves. - **Frequency** indicates how often the waves pass a point in a second.
The speed of light (\(c = 3.00 \times 10^8\, \text{m/s}\)) connects wavelength (\(\lambda\)) and frequency (\(f\)) through the equation \(c = \lambda f\). When electrons in an atom shift energy levels, they emit electromagnetic radiation as light with specific wavelengths.

Hydrogen Atom Transitions

In a hydrogen atom, electrons usually orbit the nucleus in specific energy levels or orbits. When an electron jumps from a higher orbit to a lower one, it releases energy in the form of electromagnetic radiation. This is called an electron transition.

Each transition between specific energy levels produces light of specific wavelengths. The Rydberg formula helps us determine these wavelengths:\[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]where \(n_1\) is the lower level and \(n_2\) is the higher level. As electrons move to lower energy levels (i.e., \(n=3 \rightarrow n=2\)), they emit light of a particular wavelength. Understanding these transitions in hydrogen is important for analyzing the light spectrum of stars and other celestial bodies.

Wavelength Calculation

The Rydberg formula allows us to calculate the wavelength of light emitted during a hydrogen atom transition. Let's consider the transition from \(n=3\) to \(n=2\):
\(\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\)
To find \(\lambda\) (the wavelength):- Subtract the inverse squares of the energy levels: \(\frac{1}{4} - \frac{1}{9}\).- Multiply this result by the Rydberg constant \(R_H\).- Finally, take the reciprocal of the result to find \(\lambda\).
Such calculations help identify the kind of light released, which can be visible light or not, depending on the wavelength.

Visible Light

Visible light is the small part of the electromagnetic spectrum that our eyes can detect. It ranges from approximately 400 nm (violet) to 700 nm (red).

When electrons in a hydrogen atom jump from from higher to lower energy levels, they may emit visible light.

• For example, a transition from \(n=3\) to \(n=2\) emits light with a wavelength of 656 nm, which is red.
• A transition from \(n=4\) to \(n=2\) emits light with a wavelength of 486 nm, observed as blue-green.

These radiation emissions within the visible range contribute to phenomena like stars twinkling and colorful nebulae in the universe.

Ultraviolet Radiation

Ultraviolet (UV) radiation rests just beyond the violet end of visible light. It has shorter wavelengths ranging from 10 nm to 400 nm and is not visible to the naked eye.

In the hydrogen atom, ultraviolet light is typically emitted when electrons transition to the \(n=1\) energy level, such as a transition from \(n=2 \rightarrow n=1\) which emits light at 121.60 nm.

• UV radiation carries more energy than visible light.
• It plays a critical role in natural processes like vitamin D production in humans.
• However, excessive exposure can be harmful to skin and eyes.

Studying UV radiation from celestial bodies helps astronomers learn about the compositional aspects of stellar objects.