Question

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of \(253.4 \mathrm{~nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lowerenergy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number. \()\)

Short answer

The principal quantum number of the lower energy state for the given emission spectral line of Be³⁺ can be found using the Rydberg formula for hydrogen-like ions: \[ \frac{1}{\lambda} = Z^2R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\] Given \(n_i = 5\), \(Z = 4\), and \(\lambda = 253.4 \times 10^{-9}\; m\), we can rearrange the formula to find \(n_f^2\): \[ n_f^2 = \frac{1}{\frac{1}{(5)^2} - \frac{1}{(4)^2 (1.097373 \times 10^7\;m^{-1})(253.4 \times 10^{-9}\; m)}}\] Finally, we take the square root and round to the nearest whole integer to find the principal quantum number of the lower energy state, \(n_f\): \[ n_f = \sqrt{n_f^2} \approx 4\] Thus, the principal quantum number of the lower energy state corresponding to this emission is approximately 4.

Step-by-step solution

Understand the Rydberg formula for hydrogen-like ions

The Rydberg formula for hydrogen-like ions is given by: \[ \frac{1}{\lambda} = Z^2R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\] where - λ is the wavelength of the emitted photon - \(Z\) is the nuclear charge (atomic number) - \(R_H\) is the Rydberg constant for hydrogen (\(R_H = 1.097373 \times 10^7 m^{-1}\)) - \(n_i\) is the initial principal quantum number - \(n_f\) is the final principal quantum number We are given the wavelength (λ), nuclear charge (Z), and initial principal quantum number (\(n_i\)), and we need to find the final principal quantum number (\(n_f\)).

Rearrange the formula to isolate nf²

To find the principal quantum number of the lower energy state (\(n_f\)), we will first need to isolate the \(n_f^2\) term in the Rydberg formula. Rearranging the formula, we have: \[ n_f^2 = \frac{1}{\frac{1}{n_i^2} - \frac{1}{Z^2 R_H \lambda}}\]

Plug in the given values to find nf²

Now, we plug in the given values into the formula: - \(n_i = 5\) - \(Z = 4\) - \(\lambda = 253.4 \times 10^{-9}\; m\) Calculating \(n_f^2\): \[ n_f^2 = \frac{1}{\frac{1}{(5)^2} - \frac{1}{(4)^2 (1.097373 \times 10^7\;m^{-1})(253.4 \times 10^{-9}\; m)}}\]

Calculate nf and round the value

After solving for \(n_f^2\), we will take the square root to find \(n_f\): \[ n_f = \sqrt{n_f^2}\] Always round the quantum number to the closest whole integer, as quantum numbers are always whole integers. Therefore, the principal quantum number of the lower energy state for the given emission spectral line of Be³⁺ is approximated as a whole integer.

Key concepts

Rydberg Formula

The Rydberg formula is a fundamental equation in the study of atomic physics, particularly when examining the spectral lines of elements. It was originally formulated for the hydrogen atom but can be extended to hydrogen-like ions—ions with only one electron, such as \(\mathrm{He}^{+}\), \(\mathrm{Li}^{2+}\), and in our exercise, \(\mathrm{Be}^{3+}\).

The importance of the Rydberg formula lies in its use to predict the wavelength of photons emitted or absorbed when an electron transitions between energy levels in an atom. This phenomenon underpins the creation of an atom's unique emission or absorption spectra. In its extended form for hydrogen-like ions, the formula looks like this:
\[ \frac{1}{\lambda} = Z^2R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
Here, \(\lambda\) represents the wavelength of the emitted or absorbed photon, \(Z\) is the nuclear charge equivalent to the atomic number for the ion, \(R_H\) is the Rydberg constant for hydrogen, and \(n_i\) and \(n_f\) are the initial and final principal quantum numbers, respectively. It demonstrates how the quantum numbers relate to the energy levels and therefore the energy change corresponding to a specific wavelength of light.

Hydrogen-like Ions

Hydrogen-like ions are a simplistic model for understanding more complex atomic systems. In essence, any ion that resembles a hydrogen atom, meaning it has only one electron, is termed a hydrogen-like ion. These ions are critical in the study of atomic spectra because they simplify the complex interactions found in multi-electron atoms.

For instance, in the provided exercise, \(\mathrm{Be}^{3+}\) is a hydrogen-like ion because it has shed all its electrons except one. This simplification allows us to use the Rydberg formula by incorporating the factor of \(Z\), the nuclear charge. When using the Rydberg formula for these ions, one must account for the increased attraction that the single electron feels from the nucleus, due to the larger charge. In our exercise, \(Z\) has a value of 4 for \(\mathrm{Be}^{3+}\), which will considerably influence the energy levels as compared to a neutral hydrogen atom, where \(Z\) is 1.

Emission Spectral Lines

Emission spectral lines are crucial to understanding how atoms interact with light. When an electron in an atom transitions from a higher energy level (represented by a higher principal quantum number) to a lower one, a photon is emitted. The energy of this photon corresponds to the energy difference between these two levels. The spectral lines are essentially the fingerprints of these energy transitions, appearing as distinct lines on a spectrometer's reading.

In the context of our exercise involving \(\mathrm{Be}^{3+}\), knowing the initial quantum state (with \(n=5\)) and the observed wavelength (\(253.4 \mathrm{~nm}\)), we can use the Rydberg formula to find the principal quantum number of the lower-energy state. Each line in the spectrum represents a possible transition. By calculating the principal quantum number of the lower energy state, we deepen our understanding of the atomic structure and the quantized nature of energy levels within hydrogen-like ions.