Answer
The principal quantum number of the lower-energy state corresponding to this emission is \( n_i = 4 \).
Work Step by Step
The wavelength of the emission spectral line can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{{n_f}^2} - \frac{1}{{n_i}^2} \right) \]
Where:
- \( \lambda \) = wavelength of the emitted photon
- \( R \) = Rydberg constant
- \( n_f \) = final quantum number
- \( n_i \) = initial quantum number
Given:
- \( \lambda = 253.4 \mathrm{~nm} = 253.4 \times 10^{-9} \mathrm{~m} \)
- \( n_f = 5 \)
- \( Z = 4 \) (for \( \mathrm{Be}^{3+} \))
The Rydberg constant for one-electron ions is given by:
\[ R = \frac{Z^2 R_{\infty}}{n^2} \]
Where:
- \( R_{\infty} \) = Rydberg constant for hydrogen
- \( n \) = principal quantum number
Substituting the values into the Rydberg formula and solving for \( n_i \):
\[ \frac{1}{\lambda} = \frac{Z^2 R_{\infty}}{n_f^2} - \frac{Z^2 R_{\infty}}{n_i^2} \]
\[ n_i^2 = \frac{Z^2 R_{\infty}}{ \left( \frac{1}{\lambda} + \frac{Z^2 R_{\infty}}{n_f^2} \right) } \]
Now, we can substitute the given values and solve for \( n_i \):
\[ n_i^2 = \frac{4^2 \times R_{\infty}}{ \left( \frac{1}{253.4 \times 10^{-9}} + \frac{4^2 \times R_{\infty}}{5^2} \right) } \]
Solving for \( n_i \), we find:
\[ n_i \approx 4 \]
Therefore, the principal quantum number of the lower-energy state corresponding to this emission is \( n_i = 4 \).
