Question

Calculate the wavelength (in \(\mathrm{nm}\) ) of a photon emitted by a hydrogen atom when its electron drops from the \(n=7\) state to the \(n=2\) state.

Short answer

The wavelength of the emitted photon is approximately 396.8 nm.

Step-by-step solution

Use the Rydberg Formula

The wavelength of light emitted from hydrogen transitioning between energy levels can be calculated using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( R_H \) is the Rydberg constant \( \approx 1.097 \times 10^7 \, \text{m}^{-1} \). For this problem: \( n_1 = 2 \) and \( n_2 = 7 \).

Substitute the Values into the Formula

Calculate the difference in the reciprocals of the squares of the principal quantum numbers:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{7^2} \right) \]Simplifying the expression inside the parenthesis:\[ \frac{1}{4} - \frac{1}{49} = \frac{49}{196} - \frac{4}{196} = \frac{45}{196} \]

Solve for Wavelength

Plug the simplified expression back into the equation:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{45}{196} \]Calculate:\[ \frac{1}{\lambda} = 2.520 \times 10^6 \, \text{m}^{-1} \]So:\[ \lambda = \frac{1}{2.520 \times 10^6} \approx 3.968 \times 10^{-7} \, \text{m} \]

Convert Meters to Nanometers

Since the answer should be in nanometers, convert from meters to nanometers (1 m = 1,000,000,000 nm):\[ \lambda \approx 3.968 \times 10^{-7} \, \text{m} \times 1,000,000,000 \, \frac{\text{nm}}{\text{m}} = 396.8 \, \text{nm} \]

Key concepts

Hydrogen Atom Transitions

In a hydrogen atom, electrons orbit the nucleus in specific energy levels or "shells." When an electron transitions between these energy levels, it either absorbs or emits energy in the form of a photon. This movement from one level to another results in a change in the atom's energy state.
The energy difference between the two levels dictates the characteristics of the emitted or absorbed light, such as its wavelength and frequency.

• If an electron moves from a higher energy level to a lower one, it emits a photon, releasing energy.
• Conversely, if it jumps from a lower to a higher level, it absorbs a photon, gaining energy.

The example given involves an electron in a hydrogen atom dropping from the seventh energy level ( =7") to the second ( =2"). This means the atom releases energy, which is observed as emitted light.

Photon Wavelength Calculation

The calculation of the wavelength of a photon emitted during an electron transition is accomplished using the well-known Rydberg formula. This formula helps determine the wavelength of light based on changes in an electron's energy level: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where R_H" is the Rydberg constant, valued approximately at 1.097 x 10^7 \text{m}^{-1}.
In applying the Rydberg formula, we have:

• "n_1" represents the lower energy level, in this case, \( n_1 = 2 \).
• "n_2" represents the higher energy level, here \( n_2 = 7 \).

The formula essentially calculates the inverse of the wavelength, so once you solve for \( \frac{1}{\lambda} \), you take its reciprocal to find the actual wavelength. In this scenario, the calculated wavelength lets us understand the character of the light emitted when an electron transitions.

Energy Levels

Energy levels in hydrogen atoms are quantized, meaning electrons can only occupy specific levels, each corresponding to a particular energy state. When discussing these levels, each is labeled by a principal quantum number, typically represented by \(n\).
The energy associated with a particular level can be computed using an understanding of these quantum numbers:

• Higher \(n\) values correspond to higher energy levels, which are further from the nucleus.
• Lower \(n\) values are closer to the nucleus with lower energy.

As electrons move between these levels, the energy gap between these quantized states is important. This difference decides the photon's wavelength that is either emitted or absorbed.
For example, in the exercise, the electron moves from \(n=7\) to \(n=2\), demonstrating a significant energy release due to the large gap between these two states. Understanding energy levels fosters a better grasp of the processes behind spectroscopy, allowing us to study the atomic structures.