Lyman Series of Hydrogen Spectra

The Lyman series of the hydrogen spectrum consists of a set of spectral lines that are produced when an electron transitions from higher energy levels to the ground state in a hydrogen atom. These transitions result in the emission of photons with specific wavelengths.

Wavelengths in the Lyman Series

The Lyman series is characterized by its shortest wavelength, which corresponds to the transition from the first excited state (n=2) to the ground state (n=1) of the hydrogen atom. This shortest wavelength can be calculated using the formula:

1/λ = R(1/1^2 - 1/2^2)

where λ is the wavelength, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), and n is the principal quantum number.

Given that the shortest wavelength in the Lyman series is 91.2 nm, we can substitute this value into the formula to find the value of R:

1/91.2 = R(1 - 1/4)

Simplifying the equation gives:

1/91.2 = (3/4)R

R = (4/3)(1/91.2)

R ≈ 1.097 x 10^7 m^-1

Determining the Longest Wavelength

To find the longest wavelength in the Lyman series, we need to determine the transition that corresponds to this wavelength. The longest wavelength occurs when the electron transitions from an excited state with a high principal quantum number to the ground state (n=1). This transition can be represented by the equation:

1/λ = R(1/1^2 - 1/n^2)

To find the longest wavelength, we need to find the value of n that results in the largest wavelength. As n approaches infinity, the wavelength approaches its maximum value. Substituting n = ∞ into the equation gives:

1/λ_max = R(1 - 1/∞^2)

Since 1/∞^2 is practically zero, the equation simplifies to:

1/λ_max = R

λ_max = 1/R

Substituting the value of R we calculated earlier:

λ_max = 1/(1.097 x 10^7 m^-1)

λ_max ≈ 91.2 nm

Therefore, the longest wavelength in the Lyman series is approximately 91.2 nm.

Correct Answer

According to the given options, the correct answer is option 'D' (121.6 nm).