An excited hydrogen atom emits light with a wavelength of 656.7 nm to reach the energy level for which *n* = 2. In which principal quantum level did the electron begin?


To determine the principal quantum level (\( n_i \)) from which the electron began before emitting light and moving to \( n = 2 \), we can use the Rydberg formula for the wavelengths of emitted or absorbed light in the hydrogen atom:

\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)
\]

Where:
- \( \lambda \) is the wavelength of the emitted light.
- \( R_H \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).
- \( n_f \) is the final energy level (in this case \( n_f = 2 \)).
- \( n_i \) is the initial energy level (the value we need to find).

Given:
- \( \lambda = 656.7 \, \text{nm} = 656.7 \times 10^{-9} \, \text{m} \)
- \( n_f = 2 \)

We can rearrange the Rydberg formula to solve for \( n_i \):

\[
\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n_i^2} \right)
\]

\[
\frac{1}{656.7 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{n_i^2} \right)
\]

Simplifying this:

\[
\frac{1}{656.7 \times 10^{-9}} = 2.7425 \times 10^6 \, \text{m}^{-1}
\]

\[
2.7425 \times 10^6 = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{n_i^2} \right)
\]

\[
\frac{2.7425 \times 10^6}{1.097 \times 10^7} = \frac{1}{4} - \frac{1}{n_i^2}
\]

\[
0.25 = \frac{1}{n_i^2} + 0.24998
\]

\[
\frac{1}{n_i^2} = 0.00002
\]

\[
n_i^2 \approx 9^2 \Rightarrow n_i \approx 3
\]

Thus, the electron began at the principal quantum level \( n_i = 3 \).

Would you like further details or have any questions?

### Related Questions:
1. What is the significance of the Rydberg constant in hydrogen spectra?
2. How does the energy difference between levels relate to the wavelength of emitted light?
3. Why is the \( n = 2 \) level important in the hydrogen emission spectrum?
4. How can the Balmer series be identified in the hydrogen spectrum?
5. What happens if the initial energy level \( n_i \) is higher than 3?

### Tip:
When dealing with atomic spectra, remember that the wavelength of emitted or absorbed light directly relates to the difference in energy levels of the electron transitions.

An excited hydrogen atom emits light with a wavelength of 656.7 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin?

Discover how to determine the initial principal quantum level of a hydrogen atom using the Rydberg formula. This problem involves quantum mechanics concepts and the calculation of wavelengths in atomic spectra. Ideal for students in Grades 11-12.

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