Question

A line is detected in the hydrogen spectrum at \(1880 \mathrm{nm}\). Is this line in the Balmer series? Explain.

Short answer

The answer requires calculation of \( n \) using the Balmer series formula, and then checking if it is an integer greater than 2 for it to belong to the Balmer series. The exact answer can only be determined after the calculation.

Step-by-step solution

Convert wavelength to meters

To utilize the formula of Balmer series, the given wavelength must be in meters. Convert \(1880 \, \mathrm{nm}\) to meters. 1 \, \mathrm{nm} = \(1 \times 10^{-9} \, \mathrm{m}\). Therefore, \(1880 \, \mathrm{nm} = 1880 \times 10^{-9} \, \mathrm{m} = 1.88 \times 10^{-6} \, \mathrm{m}\).

Apply the Formula of Balmer Series

Substitute \( \lambda = 1.88 \times 10^{-6} \, \mathrm{m} \) in the Balmer formula to solve for \( n \). Thus, calculate \( n \) by solving \( 1/(1.88 \times 10^{-6}) = 1.097373 \times 10^7 (1/4 - 1/n^2) \) for \( n \).

Check if n is an integer

Solve for \( n \) and check if it is an integer value. Balmer series corresponds to transitions where the electron falls to the 2nd energy level from higher levels. The value of \( n \) must be an integer higher than 2 for the wavelength to be in the Balmer series. If it is not an integer or it is less than or equals to 2, the wavelength is not in the Balmer series.

Key concepts

Balmer series

The Balmer series is a set of spectral lines of hydrogen that appear when electrons transition from higher energy levels down to the second energy level, or n=2. These transitions result in the emission of light in the visible spectrum, which is why they are often colorful.
However, not every emission line resulting from electron transitions belongs to the Balmer series. Only those where the electron ends at the second energy level do. The formula used to determine the wavelength of light emitted in the Balmer series is:

• \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \)

where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant \( (1.097373 \times 10^7 \, \text{m}^{-1}) \), and \( n \) is an integer greater than 2. If the derived \( n \) is a whole number and greater than 2, the wavelength belongs to the Balmer series.

wavelength conversion

Wavelength conversion is a necessary process to ensure calculations are consistent with formulas. Physical constants, like the Rydberg constant in the Balmer formula, use standard SI units.
When you encounter wavelengths in nanometers (nm), converting them to meters (m) is required. This is done by multiplying the value in nanometers by \(1 \times 10^{-9}\).
For example, a wavelength of 1880 nm is equivalent to \(1880 \times 10^{-9} = 1.88 \times 10^{-6} \, \text{meters}\).
This conversion is essential because it ensures the correct application of mathematical models and avoid errors in physical calculations. Always remember: 1 nm = \(1 \times 10^{-9}\) meters.

energy level transitions

Energy level transitions in atoms occur when electrons move between different orbits around the nucleus. The energy emitted or absorbed during these transitions form the hydrogen emission spectrum.
When electrons fall to a lower energy level, they emit light. The frequency and wavelength of this light depend on the specific levels involved in the transition. For instance:

• In the Balmer series, transitions end at the second energy level.
• The energy difference between levels is quantized, meaning electrons can only transition between specific energy states.

To find if a wavelength fits the Balmer series, calculate the integer \( n \) using the Balmer equation, ensuring it’s greater than 2.
This checks if the wavelength results from a transition where an electron descends to the n=2 energy level, creating visible light. If not, the line doesn’t belong in the Balmer series but may fit another series like Lyman or Paschen.