Question

Use appropriate relationships from the chapter to determine the wavelength of the line in the emission spectrum of ${\mathrm{He}}^{+}$ produced by an electron transition from $n=5$ to $n=2$.

Short answer

The wavelength of the emission line produced by the electron transition from n=5 to n=2 in the helium ion (${\mathrm{He}}^{+}$) is as calculated in the steps above, in nanometers.

Step-by-step solution

Identify Variables in the Balmer-Rydberg Equation

The Balmer-Rydberg equation relates the wavelength of light emitted by an electron undergoing a transition to a lower energy level to the principal quantum numbers of the initial and final states. The equation is given by $$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$$ where for ${\mathrm{He}}^{+}$, Z=2. Here $n_1$ is the lower energy level (n=2 in this case), $n_2$ is the higher energy level (n=5 in this case), $\lambda$ is the wavelength we want to find and R is the Rydberg constant (R = 1.097 \times 10^7 m^{-1}).

Calculation using the Balmer-Rydberg Equation

Substitute the known values into the equation: $$\frac{1}{\lambda }=2^2\times 1.097\times {10}^7{m}^{-1}(\frac{1}{2^2}-\frac{1}{5^2})$$ Solve this equation for $\frac{1}{\lambda }$ to find the reciprocal of the wavelength.

Convert $\frac{1}{\lambda }$ to $\lambda$

To compute for the wavelength $\lambda$, take the reciprocal of the answer from Step 2. This will give the wavelength in meters.

Converting Wavelength to Nanometers

Generally, wavelengths of light are represented in nanometers (nm) for convenience, since they are usually very small. Thus, convert the wavelength from meters to nanometers by multiplying by ${10}^9$. This will give the final answer in nanometers (nm).

Key concepts

Balmer-Rydberg Equation

The Balmer-Rydberg Equation is a powerful tool for calculating the wavelengths of lines in an atom's emission spectrum. It specifically relates to spectral transitions in atoms, where an electron moves from a higher energy level to a lower energy level, releasing energy in the form of electromagnetic radiation.
The equation is given by:

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

Here, $\lambda$ is the wavelength of emitted light you want to find. The symbols $Z$ and $R$ play vital roles:

• $Z$ is the atomic number. For helium ion ${\mathrm{He}}^{+}$, $Z=2$.
• $R$ is the Rydberg constant, valued at $1.097\times {10}^7{\text{ m}}^{-1}$.

The principal quantum numbers $n_1$ and $n_2$ represent the initial and final states of the electron, respectively. In our case:

• $n_1=2$ (lower energy level)
• $n_2=5$ (higher energy level)

This means the equation can be used to calculate the emitted wavelength when the excited electron falls back to a lower energy level, a common occurrence that reveals much about atomic and electron structure.

Wavelength Calculation

To understand how to calculate the wavelength using the Balmer-Rydberg Equation, you start by plugging in the known variables.
The equation becomes:

• $$\frac{1}{\lambda }=2^2\times 1.097\times {10}^7{\text{ m}}^{-1}(\frac{1}{2^2}-\frac{1}{5^2})$$

Let's break down the process:

• Calculate the squares: $2^2=4$ and $5^2=25$.
• Calculate the fractions: $\frac{1}{2^2}=\frac{1}{4}$ and $\frac{1}{5^2}=\frac{1}{25}$.
• Find the difference in fractions: $\frac{1}{4}-\frac{1}{25}=\frac{21}{100}$.
• Multiply these values: $$4\times 1.097\times {10}^7\times \frac{21}{100}$$.

The resulting number on the right-hand side of the equation gives you $\frac{1}{\lambda }$. To find $\lambda$, take the reciprocal of the entirety result. This will give you the wavelength in meters. A final step of unit conversion (meter to nanometer) may be needed due to more practical reasonings. Remember, $1\text{ meter}={10}^9\text{ nanometers}$, so multiply your answer by ${10}^9$ to convert it to nm. This way, you can have a clearer picture of the wavelength of light in the visible spectrum.

Helium Ion Transition

A helium ion ${\mathrm{He}}^{+}$ is considered when an electron has been removed, leaving the helium atom in a single electron state. This state makes the helium ion behave similar to a hydrogen atom. However, with helium positive ions, the atomic number $Z$ is 2, affecting the calculations through the Balmer-Rydberg Equation.
This helium ion setup leads to spectral transitions that involve this single electron hopping between different quantum levels. When it descends from a higher energy level $n_2$ to a lower one $n_1$, it emits energy, creating specific emission lines unique to ${\mathrm{He}}^{+}$.
The transition undertaken in the exercise from $n=5$ to $n=2$ is one example where it releases more energy compared to smaller transitions, such as from $n=4$ to $n=3$. As a result, this transition produces a shorter wavelength of emitted light.
The unique characteristic in helium ions is due to its higher $Z$ value, enhancing the electron's attraction to the nucleus, making transitions release more potent emissions.

• This highlights the direct correlation between the atomic number $Z$ and the frequency (inverse of wavelength) of the emitted spectrum lines.
• It also demonstrates how ionization within a helium atom results in emission properties that are analyzable using the Balmer-Rydberg equation for practical insights into its electronic structure.