Shortest and Longest Wavelength in Li2 Spectrum of Balmer Series

To calculate the shortest and longest wavelength in the Li2 spectrum of the Balmer series, we need to understand the Balmer series and its relationship with the energy levels of the Li2 atom.

Balmer Series
The Balmer series is a series of spectral lines in the visible region of the electromagnetic spectrum that corresponds to transitions between the energy levels of a hydrogen-like atom. It was first discovered by Johann Balmer in 1885 and is expressed by the Balmer formula:

1/λ = R_H * (1/n1^2 - 1/n2^2)

Where:
- λ is the wavelength of the spectral line
- R_H is the Rydberg constant for hydrogen
- n1 and n2 are the principal quantum numbers of the energy levels involved in the transition

Lithium (Li2) Atom
The Li2 atom consists of two lithium atoms that are chemically bonded together, forming a diatomic molecule. The electron configuration of a lithium atom is 1s^2 2s^1, meaning it has two electrons in the innermost 1s orbital and one electron in the 2s orbital.

Calculating Shortest Wavelength
To determine the shortest wavelength in the Li2 spectrum of the Balmer series, we need to find the transition that involves the highest energy levels. In the case of Li2, the highest energy level available for transitions is the 3s orbital. Therefore, we can use the Balmer formula to calculate the wavelength for the transition from n1 = 3 to n2 = 2.

1/λ = R_H * (1/2^2 - 1/3^2)

Simplifying the equation gives:

1/λ = R_H * (1/4 - 1/9)

1/λ = R_H * (5/36)

λ = 36/(5 * R_H)

The value of the Rydberg constant for hydrogen (R_H) is approximately 1.097 x 10^7 m^-1. Substituting this value into the equation:

λ = 36/(5 * 1.097 x 10^7)

λ ≈ 6.55 x 10^-7 m

Therefore, the shortest wavelength in the Li2 spectrum of the Balmer series is approximately 6.55 x 10^-7 meters, which corresponds to the transition from the 3s to the 2s energy level.

Calculating Longest Wavelength
To determine the longest wavelength in the Li2 spectrum of the Balmer series, we need to find the transition that involves the lowest energy levels. In the case of Li2, the lowest energy level available for transitions is the ground state, which is the 2s orbital. Therefore, we can use the Balmer formula to calculate the wavelength for the transition from n1 = 2 to n2 = 3.

1/λ = R_H * (1/3^2 - 1/2^2)

Simplifying the equation gives:

1/λ = R_H * (1/9 - 1/4)

1/λ = R_H * (4/36 - 9/36)

1/λ = R_H * (-5