In Li++, an electron in the first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ?

The number of spectral lines emitted when an electron de-excites from a higher energy level (n) to the ground state (n=1) is given by n(n-1)/2. Since six spectral lines are observed, we have:

n(n-1)/2 = 6
n(n-1) = 12
n² - n - 12 = 0
(n-4)(n+3) = 0
Since n must be positive, n = 4.

This means the electron was excited from n=1 to n=4.
The energy difference between these levels in a hydrogen-like ion (Li²⁺) is given by:

ΔE = -13.6 Z²(1/n₂² - 1/n₁²) eV

where Z is the atomic number (Z=3 for Li), n₁ is the initial energy level (n₁=1), and n₂ is the final energy level (n₂=4).

ΔE = -13.6 * 3²(1/4² - 1/1²) = -13.6 * 9 * (1/16 - 1) = -13.6 * 9 * (-15/16) = 114.75 eV

The energy of the photon absorbed is equal to the energy difference:

ΔE = hc/λ

where h is Planck's constant (4.136 x 10⁻¹⁵ eV·s), c is the speed of light (3 x 10⁸ m/s), and λ is the wavelength.

Solving for λ:

λ = hc/ΔE = (4.136 x 10⁻¹⁵ eV·s * 3 x 10⁸ m/s) / 114.75 eV ≈ 1.08 x 10⁻⁸ m = 10.8 nm

Therefore, the value of λ is 10.8 nm.