The Balmer series corresponds to the spectral lines emitted by hydrogen atoms when electrons transition from higher energy levels to the second energy level (n = 2). The wavelength of the first line in the Balmer series, known as the Balmer alpha line, is indeed approximately 656.3 nm.
To calculate the energy difference between the two energy levels involved in this emission, we can use the Rydberg formula:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2),
where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1), and n_initial and n_final are the principal quantum numbers representing the initial and final energy levels, respectively.
In this case, the initial energy level is higher, so n_initial is greater than n_final. For the Balmer alpha line, n_initial is 3 (transition from n = 3 to n = 2) and n_final is 2.
Let's calculate the energy difference:
1/λ = R_H * (1/2^2 - 1/3^2).
1/λ = R_H * (1/4 - 1/9).
1/λ = R_H * (9/36 - 4/36).
1/λ = R_H * (5/36).
λ = 36/5R_H.
Now we can substitute the value of R_H, which is approximately 1.097 x 10^7 m^-1, and calculate the value of λ:
λ = (36/5) / (1.097 x 10^7) m.
To convert this value to nanometers, we multiply by 10^9:
λ = (36/5) / (1.097 x 10^7) x 10^9 nm.
Calculating this expression, we find:
λ ≈ 656.3 nm.
Therefore, the energy difference between the two energy levels involved in the emission that results in the Balmer alpha line is given by the wavelength of the line, which is approximately 656.3 nm.