To calculate the wavelength of the photon emitted during an electron transition from $n = 6$ to $n = 3$ in a hydrogen atom, we can use the Rydberg formula:

$\frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$

where:

• $\lambda$ is the wavelength of the emitted photon.
• $R$ is the Rydberg constant, $R = 1.097 \times 10^7 \, \text{m}^{-1}$.
• $n_1$ and $n_2$ are the initial and final energy levels of the electron, respectively.

Given:

• $n_1 = 6$
• $n_2 = 3$

We first calculate the frequency of the photon using the energy difference, then convert it to wavelength using the speed of light equation $c = \lambda \nu$.

Let's plug in the values and compute the wavelength.The wavelength of the photon emitted during the transition from $n = 6$ to $n = 3$ is approximately 1093.9 nm (nanometers).

Would you like more details or have any questions? Here are five related questions to explore further:

1. What is the significance of the Rydberg constant in quantum mechanics?
2. How does the energy difference between electron levels relate to photon emission?
3. Can this wavelength fall within the visible spectrum?
4. How would the wavelength change if the electron transitioned from $n = 5$ to $n = 3$?
5. What is the relationship between photon wavelength and energy?

Tip: The energy of a photon is inversely proportional to its wavelength, meaning shorter wavelengths correspond to higher energy photons.