Question

When an electron makes a transition from the \(n=3\) to the \(n=2\) hydrogen atom Bohr orbit, the energy difference between these two orbits \(\left(3.0 \times 10^{-19} \mathrm{~J}\right)\) is emitted as a photon of light. The relationship between the energy of a photon and its wavelength is given by \(E=h c / \lambda\), where \(E\) is the energy of the photon in \(J, h\) is Planck's constant \(\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\), and \(c\) is the speed of light \(\left(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\). Find the wavelength of light emitted by hydrogen atoms when an electron makes this transition.

Short answer

The wavelength of light emitted is \(6.626 \times 10^{-7} \mathrm{~m}\) or \(662.6\) nm.

Step-by-step solution

Identify Given Information

First, note the given values: the energy difference between the orbits is given as \(3.0 \times 10^{-19} \mathrm{~J}\), Planck's constant \(h\) is \(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\), and the speed of light \(c\) is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). You are asked to find the wavelength \(\lambda\).

Use the Energy-Photon Relationship

Use the equation relating the energy of a photon to its wavelength \(E = h c / \lambda\) to find the wavelength \(\lambda\).

Solve for the Wavelength \(\lambda\)

Rearrange the equation to solve for \(\lambda\): \(\lambda = h c / E\). Substitute in the values to find \(\lambda\): $$\lambda = \frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} \times 3.00 \times 10^{8} \mathrm{~m/s}}{3.0 \times 10^{-19} \mathrm{~J}}.$$ Simplify to find the numerical value for \(\lambda\).

Calculate the Wavelength

Perform the actual calculation: $$\lambda = \frac{(6.626 \times 10^{-34}) \times (3.00 \times 10^{8})}{3.0 \times 10^{-19}} = \frac{1.9878 \times 10^{-25}}{3.0 \times 10^{-19}} = 6.626 \times 10^{-7} \mathrm{~m}$$ Therefore, the wavelength of the emitted light is \(6.626 \times 10^{-7}\) meters, which can also be written as \(662.6\) nm, since \(1\) m \(= 10^{9}\) nm.

Key concepts

Electron Transitions in Hydrogen

Understanding electron transitions in a hydrogen atom is foundational for studying quantum mechanics and atomic physics. The hydrogen atom, the simplest of atoms, has a nucleus consisting of a single proton around which a single electron moves. Niels Bohr proposed that an electron in a hydrogen atom occupies specific energy levels, or orbits, without emitting radiation. These discrete orbits are indexed by quantum numbers, labelled as n, where n = 1, 2, 3, ... and correspond to increasing energy levels.

When an electron jumps from a higher energy level to a lower one, it emits energy in the form of a photon, a particle of light. The transition mentioned in the exercise, from the third (n=3) to the second (n=2) energy level, is just one example of how an electron can move between these quantized orbits. The energy difference between these levels precisely matches the energy carried away by the emitted photon. This phenomenon is not only crucial to understand atomic structure but also has applications in technologies such as lasers and spectral analysis.

Providing Context for the n-Values

• The n=3 level is known as the third Bohr orbit, which is a higher energy state compared to the n=2 level, the second Bohr orbit.
• Each transition has a unique energy value which quantifies the photon emitted during the transition; this energy is the 'energy difference' referring to the exercise.

Photon Energy and Wavelength Relationship

The relationship between the energy of a photon and its wavelength is a core concept in the study of light and electromagnetism. The energy (E) of a photon is inversely proportional to its wavelength (λ), which is expressed using the equation E = hc/λ, where h is Planck's constant and c is the speed of light in vacuum. This fundamental relationship implies that photons with shorter wavelengths have higher energies, and those with longer wavelengths have lower energies.

The equation E = hc/λ can also be rearranged to solve for the wavelength of the photon when its energy is known, which was the task in our exercise. The equation is reconfigured into λ = hc/E to find the wavelength associated with an energy released during an electron transition. It's important to understand that this equation describes the theoretical limit of the energy-wavelength relationship. In practical applications, factors such as the medium through which the light is traveling can affect the actual speed of light and, consequently, the wavelength measurements.

Connecting Energy and Color

• Visible light spectrum ranges from about 400 nm (violet) to 700 nm (red); transitions releasing photons within this range will emit visible light.
• The transition described in the exercise results in a photon with a wavelength in the visible range, specifically around 662.6 nm, which corresponds to red-orange light.

Planck's Constant Application

Planck's constant (h) is a fundamental quantity in quantum mechanics, denoting the smallest action in nature and linking the amount of energy a photon carries with its frequency. Its application is pivotal in determining the characteristics of photons emitted or absorbed during the electron transitions in atoms. In the exercise, Planck's constant is used along with the speed of light (c) to calculate the wavelength of light emitted when an electron transitions between energy levels in the hydrogen atom.

In technical terms, Planck's constant provides the scale of quantum effects. Without it, calculations for energy levels in quantum systems could not be performed, and the discrete nature of energy in atomic systems would remain unexplained. Moreover, its value of approximately 6.626 x 10^-34 Js is essential in calculations like the one worked through in the exercise, where it is crucial for finding the exact wavelength of the emitted photon during the electron transition.

The Significance of Planck's Constant

• The precise value of h ensures accurate calculations of energy and wavelength in quantum phenomena.
• Understanding its application is essential for interpreting spectral lines, which are used to identify elements in stars and distant galaxies.