Final answer:
The shortest wavelength of the Balmer series is four times that of the Lyman series, indicating that the Balmer series wavelengths are longer and lie in the visible range, while the Lyman series is in the ultraviolet range.
Step-by-step explanation:
The question asks about the shortest wavelengths for the Balmer and Lyman series in a hydrogen atom spectrum. Using the Rydberg formula, the shortest wavelength for the Balmer series (λ_Balmer) occurs when the electron transition ending level is n=2 and the starting level approaches infinity (n∞). For the Lyman series, the shortest wavelength (λ_Lyman) occurs when the electron transition ends at n=1 and starts again at infinity (n∞). The Rydberg constant (R) is 1.097 x 107 m−1.
The ratio of these wavelengths can be found using the Rydberg formula: 1/λ = R (1/nf2 - 1/ni2). For the shortest wavelength in the Balmer series, nf=2 and ni=∞, giving us 1/λ_Balmer = R (1/22 - 1/∞2). Simplifying this, we get 1/λ_Balmer = R/4.
For the shortest wavelength in the Lyman series, nf=1 and ni=∞, leading to 1/λ_Lyman = R (1/12 - 1/∞2). This simplifies to 1/λ_Lyman = R.
Now, the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is λ_Lyman / λ_Balmer = (R/4) / R = 1 / 4. Therefore, the shortest wavelength of the Balmer series is four times that of the Lyman series. This also shows that the shortest wavelength of the Balmer series is longer compared to the Lyman series, and thus, they lie in different regions of the electromagnetic spectrum with Balmer series in the visible range and the Lyman series in the ultraviolet (UV) range.