Hydrogen exhibits several series of line spectra in different spectral regions. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the visible range and the Paschen (nf = 3), Brackett (nf = 4) and Pfund ( nf = 5) series all occur in the infrared range. What is the shortest-wavelength (in nm) in the Balmer series?
See your post on the Brackett series. Follow the same guidelines.
To find the shortest-wavelength in the Balmer series, we need to use the Balmer-Rydberg equation. This equation relates the wavelength of the spectral lines emitted by hydrogen to the energy levels of the electrons in the atom.
The Balmer-Rydberg equation is given by:
1/λ = R * (1/nf^2 - 1/ni^2)
Where:
λ is the wavelength of the spectral line,
R is the Rydberg constant (1.097373 x 10^7 m^-1),
nf is the final energy level of the electron, and
ni is the initial energy level of the electron.
In the Balmer series, nf is always 2 because it is the visible range. The initial energy levels ni for the different lines in the Balmer series can be found by substituting nf = 2 into the equation:
1/λ = R * (1/2^2 - 1/ni^2)
1/λ = R * (1/4 - 1/ni^2)
To find the shortest-wavelength, we need to find the highest value of ni. Since nf = 2 and ni > nf, we can start with ni = 3:
1/λ = R * (1/4 - 1/3^2)
1/λ = R * (1/4 - 1/9)
1/λ = R * (5/36)
To find the value of λ, we can rearrange the equation:
λ = 36/(5R)
Plugging in the value of the Rydberg constant, we get:
λ = 36/(5 * 1.097373 x 10^7)
Calculating this equation gives us the wavelength in meters. However, the question asks for the shortest wavelength in nanometers. To convert meters to nanometers, we multiply by 10^9:
λ = (36/(5 * 1.097373 x 10^7)) * 10^9
Evaluating this expression gives us the shortest-wavelength in the Balmer series in nanometers.