The value of λ C. 10.8 nm .Therefore , C. 10.8 nm is correct.
To determine the value of λ, we need to consider the energy levels of the Li++ ion and the number of spectral lines observed during deexcitation.
The energy levels of a hydrogen-like ion (such as Li++) are given by the Rydberg formula:
E = -13.6 eV/n^2
where n is the principal quantum number (n = 1 for the ground state).
When the electron is excited to a higher energy level, it absorbs energy in the form of a photon with wavelength λ.
The energy of this photon can be calculated using the following equation:
E = hc/λ
where h is Planck's constant (6.63 x 10^-34 Js) and c is the speed of light (3 x 10^8 m/s).
Since the electron can deexcite to the ground state in all possible ways, there will be a total of 6 spectral lines.
This means that there are 6 different energy transitions between the excited state and the ground state.
To determine the value of λ, we need to know the energy difference between the excited state and the ground state.
Let's assume that the excited state corresponds to n = 4.
The energy difference between n = 4 and n = 1 is:
ΔE = -13.6 eV (1/1^2 - 1/4^2) = -3.4 eV
Now we can use the equation E = hc/λ to solve for λ:
3.4 eV = (6.63 x 10^-34 Js)(3 x 10^8 m/s)/λ
λ = (6.63 x 10^-34 Js)(3 x 10^8 m/s)/(3.4 eV x 1.6 x 10^-19 J/eV)
λ = 10.8 nm
Therefore, the value of λ is 10.8 nm. The answer is C. 10.8 nm.