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 lower-energy 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 corresponding to this emission is 4.

Step-by-step solution

Calculate the Energy of Emission Spectral Line

First, we will calculate the energy of the emission spectral line using the given wavelength. The energy of a photon is given by: $$E = h \cdot c / \lambda$$ Where: - \(\lambda\) is the wavelength of the emission spectral line - \(E\) is the energy - \(h\) is the Planck's constant (\(6.63 \times10^{-34} \ \mathrm{J \cdot s}\) ) - \(c\) is the speed of light (\(3.00 \times10^{8} \ \mathrm{m/s}\)) First, convert the given wavelength from nanometers to meters: $$\lambda = 253.4 \times10^{-9} \ \mathrm{m}$$ Then, calculate the energy: $$E = \frac{6.63 \times10^{-34} \ \mathrm{J \cdot s} \cdot 3.00 \times10^{8} \ \mathrm{m/s}}{253.4 \times10^{-9} \ \mathrm{m}}$$ $$E = 7.87 \times10^{-19} \ \mathrm{J}$$

Calculate the Principal Quantum Number of Lower-Energy State

Now, we will use the Rydberg formula to solve for the principal quantum number of the lower-energy state. Rearranging the equation so that \(n^2\) is the subject, we have: $$n^2 = \frac{hcRZ^2}{E}$$ Where: - \(E\) is the change in energy from Step 1 (\(7.87 \times10^{-19}\ \mathrm{J}\)) - \(Z\) is the nuclear charge (4 for \(\mathrm{Be}^{3+}\)) - \(R\) is the Rydberg constant for hydrogen (\(1.097 \times10^{7} \ \mathrm{m^{-1}}\)) We need to find \(n^2\) and then take the square root of the answer to find the final principal quantum number, \(n\). We have: $$n^2 = \frac{(6.63 \times10^{-34}\ \mathrm{J \cdot s}) \cdot(3.00 \times10^{8} \ \mathrm{m/s}) \cdot (1.097 \times10^{7} \ \mathrm{m^{-1}}) \cdot (4)^2}{7.87 \times10^{-19} \ \mathrm{J}}$$ $$n^2 \approx 20.52$$ Now, take the square root to determine the principal quantum number, rounding to the nearest whole number: $$n = \sqrt{20.52} \approx 4.5$$ Since the quantum number must be a whole number, we round it up to the nearest integer, \(n = 5\). However, this would mean that the initial and final quantum number would be the same, which is not possible for an electronic transition. Therefore, we round it down to the nearest integer, \(n=4\). The principal quantum number of the lower-energy state corresponding to this emission is 4.

Key concepts

Emission Spectral Line

An emission spectral line is a specific wavelength of light that is emitted when an electron in an atom transitions from a higher energy level to a lower one. When electrons drop from a higher orbit to a lower one, they release energy in the form of light, which we see as a spectral line.

To imagine how this works, think of the electron as a ball on a set of steps. When you allow the ball to drop from a higher step to a lower one, it releases energy. In atoms, this "energy drop" occurs as an emission of light. The specific characteristics of this light depend heavily on the principal quantum numbers of the involved energy levels.

Emission lines can be incredibly useful in understanding atomic structures. For instance, the wavelengths and frequencies of lines can tell us about the energy differences between these levels. It is an essential concept for analyzing atomic and molecular compositions in various fields including astrophysics and chemistry.

Principal Quantum Number

The principal quantum number, denoted as \(n\), is one of the key numbers used to describe the electron’s state in an atom. Think of it as the address of an electron's location within an atom, determining the energy level or shell in which the electron resides. They are integral values, typically starting from 1 (but can be as high as the electron allows).

The higher the number \(n\), the farther the electron is from the nucleus and the higher its energy. When an electron moves between energy levels with different principal quantum numbers, it results in either absorption or emission of energy, visible as a spectral line.

• The energy of an electron in a particular level is usually denoted as \(E_n\).
• Changes in \(n\) are what lead to the phenomena such as spectral emissions.

Each element can be thought of having its own unique set of energy levels and transitions, often explored through the Bohr model.

Rydberg Formula

The Rydberg formula is a mathematical equation used in atomic physics to predict the wavelengths of spectral lines. It is particularly applicable for hydrogen-like atoms, which include any ion with only one electron, like \(\mathrm{Be}^{3+}\). This powerful tool helps to determine the wavelengths of light emitted or absorbed during an electron's transition between energy levels.

The formula is expressed as: \[ \frac{1}{\lambda} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]Where:

• \(\lambda\) is the wavelength of the emitted light.
• \(R\) is the Rydberg constant (approximated as \(1.097 \times 10^7 \ \mathrm{m^{-1}}\) for hydrogen).
• \(Z\) is the atomic number, representing nuclear charge.
• \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy states respectively.

By inputting known variables into this formula, scientists can derive the characteristics of light emissions. It’s an essential aspect for understanding the atomic spectrum and plays a significant role in studying atomic interactions in various scientific fields.