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 150 .

Short answer

The color of light corresponding to the electronic transition from n=6 to n=2 in the hydrogen emission spectrum is violet. This result is obtained by calculating the wavelength of the light emitted during the transition using the Rydberg formula, which is found to be approximately \(410.3 \mathrm{nm}\), and looking up the color corresponding to this wavelength.

Step-by-step solution

Understanding the Rydberg formula

The Rydberg formula is used to calculate the wavelength of the light emitted due to electronic transitions in the hydrogen atom. The formula is given by: \( \dfrac{1}{\lambda} = R_{H} \left(\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right) \) Where: - \( \lambda \) is the wavelength of the light emitted, - \( R_{H} \) is the Rydberg constant for hydrogen, approximately equal to \( 1.097 \times 10^{7} \mathrm{m}^{-1} \), - \( n_{1} \) and \( n_{2} \) are the initial and final energy levels, respectively. In this case, we have the transitions from \( n_{1} = 6 \) to \( n_{2} = 2 \).

Calculate the wavelength

Using the Rydberg formula, we can calculate the wavelength of the light emitted during the transition from n=6 to n=2. \( \dfrac{1}{\lambda} = R_{H} \left(\dfrac{1}{2^{2}} - \dfrac{1}{6^{2}}\right) \) |Plug in the values and solve for \( \lambda \). \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \left(\dfrac{1}{4} - \dfrac{1}{36}\right) \) \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \left(\dfrac{32}{144}\right) \) \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \times \dfrac{8}{36} \) Now, find \( \lambda \) by taking the inverse: \( \lambda = \dfrac{36}{1.097 \times 10^{7} \mathrm{m}^{-1} \times 8} \) \( \lambda \approx 4.103 \times 10^{-7} \mathrm{m} \) \( \lambda \approx 410.3 \mathrm{nm} \)

Determine the color of the light

The color of the light can be determined by looking up the wavelength in a table of visible light colors or using an online tool. The wavelength of \( 410.3 \mathrm{nm} \) corresponds to the color violet, which lies around spectral range (400nm - 450nm). Therefore, the color of the light emitted during the transition from n=6 to n=2 is violet.

Key concepts

Hydrogen Emission Spectrum

The hydrogen emission spectrum is a key topic in understanding atomic structure and quantum mechanics. When an electron in a hydrogen atom transitions between different energy levels, it either absorbs or emits energy in the form of light. This process results in the emission of light at specific wavelengths, forming a series of lines known as the hydrogen emission spectrum. Each line in this spectrum corresponds to a distinct electronic transition.

The spectrum is divided into different series, each named after the scientist who discovered them. The Balmer series, for instance, appears in the visible spectrum and is particularly notable because these lines are visible to the human eye. The exercise at hand involves a transition in the hydrogen atom from the sixth energy level ( =6 e) to the second energy level ( =2 e), specifically within the Balmer series. This results in the emission of visible light, allowing us to directly observe the effects of quantum mechanics at work.

Electronic Transitions

Electronic transitions in an atom occur when an electron moves from one energy level to another. These transitions can involve the absorption or emission of light, depending on the direction of the change. When an electron falls from a higher energy level to a lower one, as from =6 e to =2 e in hydrogen, energy is released, and light is emitted.

The Rydberg formula is pivotal in calculating the wavelengths of light associated with these transitions. The formula considers the initial and final energy levels to determine the wavelength of the emitted light. Through calculations using the Rydberg formula, it's possible to identify the nature of the light emitted, such as the color which corresponds to specific wavelengths within the electromagnetic spectrum.

Visible Light Spectrum

The visible light spectrum comprises the portion of the electromagnetic spectrum that is visible to the human eye. It ranges from approximately 400 nm, with violet light, to about 700 nm, with red light. Different wavelengths within this range correspond to various colors that we perceive.

In the context of the hydrogen emission spectrum, when the electron transition from =6 e to =2 e occurs, a wavelength of roughly 410.3 nm is emitted. This places the light in the violet range of the visible spectrum.

• Wavelengths between 400 nm and 450 nm are typically perceived as violet or purple.
• Understanding these wavelengths helps in identifying the color of the light emitted during electronic transitions in hydrogen.

By observing where these wavelengths fall within the visible spectrum, scientists can better understand the properties and behaviors of atoms under different conditions.