Two of the emission wavelengths in the hydrogen emission spectrum are \(656 \mathrm{~nm}\) and \(486 \mathrm{~nm}\). One of these is due to the \(n=4\) to \(n=2\) transition, and the other is due to the \(n=3\) to \(n=2\) transition. Which wavelength corresponds to which transition?

When it comes to analyzing the hydrogen emission spectrum, the Rydberg formula is an essential tool. This formula is used to predict the wavelengths of light that a hydrogen atom will emit when an electron makes a transition between energy levels.

The Rydberg formula is often stated as:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:\

• \( \lambda \) is the wavelength of the emitted light,
• \( R \) represents the Rydberg constant, which is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \),
• \( n_1 \) is the lower energy level (final state),
• and \( n_2 \) is the higher energy level (initial state).

Understanding and applying this formula allows us to solve problems involving the spectral lines of hydrogen and can aid in grasping the broader concept of atomic emission in quantum physics.

Atomic emission occurs when an atom's electron transitions from a higher energy level to a lower one, releasing a photon of light in the process. This light can be observed as a line in the emission spectrum, which is a series of discrete wavelengths. The hydrogen emission spectrum is particularly simple and well-defined, making it an excellent subject to study the principles of atomic emission.

Each line in the emission spectrum corresponds to a different electron transition. For an atom like hydrogen with one electron, the energy levels are well understood, and the transitions can be predicted using the Rydberg formula. Specifically, when an electron drops from a higher orbit (like from the third to the second orbit), it emits a photon with a specific wavelength, contributing to the unique pattern observed in the hydrogen spectrum.

An electron transition refers to the movement of an electron between different energy levels within an atom. As quantized by Niels Bohr and others, these energy levels are labeled by an integer value known as the principal quantum number \( n \). Transitions can either absorb energy (moving to a higher level, or 'excited state') or release energy (returning to a lower level, or 'ground state') in the form of light (photons).

The particular energy difference between the levels determines the wavelength of the emitted photon. In the step-by-step solution, transitions from \( n=4 \) to \( n=2 \) and from \( n=3 \) to \( n=2 \) are explored, showing how the differences in these transitions result in distinct wavelengths of light that, which are roughly \( 486 \, \mathrm{nm} \) and \( 656 \, \mathrm{nm} \), respectively.

The wavelength calculation is done by applying the known values into the Rydberg formula. After you've determined the initial and final energy levels of the electron transition, these values are substituted into the formula to solve for the wavelength of light that corresponds to that particular transition.

For example, the calculation for the \( n=4 \) to \( n=2 \) transition would look like this:
\[ \lambda = \frac{1}{R \left( \frac{3}{16} \right)} = \frac{16}{3R} \]
Similarly, the calculation for the \( n=3 \) to \( n=2 \) transition would be:
\[ \lambda = \frac{1}{R \left( \frac{5}{36} \right)} = \frac{36}{5R} \]
Comparing the calculated wavelengths to known spectral lines allows us to correlate transitions to specific lines within the hydrogen emission spectrum.