Problem 21 - College Physics

Question

An electron in an excited state of hydrogen makes a transition from the \(n=5\) level to the \(n=2\) level. (a) Does the atom emit or absorb a photon during this process? How do you know? (b) Calculate the wavelength of the photon involved in the transition.

Step-by-Step Solution

Answer

(a) The atom emits a photon. (b) The wavelength of the photon is approximately 434 nm.

Step 1: Determine Emission or Absorption

An electron in an atom transitions from a higher energy level to a lower one, it emits a photon. Since the electron is moving from \(n=5\) to \(n=2\), which is a higher to lower energy transition, it emits a photon.

Step 2: Use the Rydberg Formula

To find the wavelength of the emitted photon, use 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 \(1.097 \times 10^7 \text{ m}^{-1}\), \(n_1 = 2\), and \(n_2 = 5\). Insert these values into the formula.

Step 3: Calculate the Term Values

First, calculate \( \frac{1}{n_1^2} = \frac{1}{2^2} = \frac{1}{4} = 0.25 \) and \( \frac{1}{n_2^2} = \frac{1}{5^2} = \frac{1}{25} = 0.04 \).

Step 4: Calculate the Difference and Plug Into Rydberg Formula

Subtract the two term values obtained: \( 0.25 - 0.04 = 0.21 \).Plug this difference into the Rydberg formula:\[\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.21\].

Step 5: Solve for Wavelength \( \lambda \)

Calculate the reciprocal value to determine the wavelength:\[\frac{1}{\lambda} = 2.3037 \times 10^6 \text{ m}^{-1}\]This means \( \lambda \approx \frac{1}{2.3037 \times 10^6} = 4.34 \times 10^{-7} \text{ meters} \) or \(434\text{ nm}\).

Key Concepts

Photon Emission

When an electron transitions between different energy levels within an atom, it undergoes a process known as photon emission if it moves to a lower energy level. This occurs because the electron loses energy, and that energy is released in the form of a photon.
In the exercise with hydrogen, as the electron transitions from the higher energy level of \(n=5\) to \(n=2\), it emits a photon.

• Photon emission always results from a transition down to a lower energy level.
• The energy difference between these levels determines the energy and thus the properties of the emitted photon.

Understanding photon emission is crucial for grasping how light is generated at the atomic level and for exploring spectral lines in physics and chemistry.

Rydberg Formula

The Rydberg formula is an essential tool for calculating the wavelengths of photons emitted or absorbed by electrons moving between energy levels in hydrogen-like atoms. The formula is expressed by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where:

• \(\lambda\) is the wavelength of the photon,
• \(R_H\) is the Rydberg constant \(1.097 \times 10^7 \text{ m}^{-1}\),
• \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy levels, respectively.

This formula is extremely helpful in determining the specific wavelengths of light associated with transitions in hydrogen, the simplest element.

Energy Levels

Energy levels in an atom are like the steps of an invisible staircase for electrons. Each step represents a distinct quantum state, often labeled with quantum numbers such as \(n\).
The levels closest to the nucleus are the lowest in energy. When electrons jump between these levels, the atom absorbs or emits energy as a photon, depending on whether the transition is upward or downward.

• A transition from high to low energy levels emits a photon.
• A transition from low to high energy levels absorbs a photon.

In our exercise, an electron transitioned from \(n=5\) to \(n=2\), moving from a higher to a lower energy state, resulting in photon emission.

Wavelength Calculation

Calculating the wavelength of the emitted photon involves putting all the pieces together using the Rydberg formula. In our specific case, as the electron moves from \(n=5\) to \(n=2\), we start by finding the initial and final term values:

• First, \( \frac{1}{n_1^2} = \frac{1}{2^2} = \frac{1}{4} = 0.25 \)
• Second, \( \frac{1}{n_2^2} = \frac{1}{5^2} = \frac{1}{25} = 0.04 \)

Then subtract these values: \( 0.25 - 0.04 = 0.21 \).
Plugging this difference into the Rydberg formula gives:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.21 \]
When calculating the reciprocal, the resulting wavelength is approximately \(434 \text{ nm}\), which corresponds to visible light within the violet-blue part of the spectrum.