The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is x × 10⁻⁸ m. The value of x is ___________. (Nearest Integer).
(Given : Energy of the electron in the first shell of the hydrogen atom is -2.2 × 10⁻¹⁸ J ; h = 6.63 × 10⁻³⁴ Js and c = 3 × 10⁸ ms⁻¹)

Solution

Chemistry ma’am · 4 steps

Lithium's outermost electron sits in n = 2, not n = 1, so put n = 2 and Z = 3 into the Bohr energy formula.

1. Which electron leaves. Lithium is 1s² 2s¹, so ionisation removes the electron in the second shell. The longest wavelength that can ionise it carries exactly the energy needed to take this electron from n = 2 to n = ∞.

   Eₙ = E₁(H) × Z²/n² = -2.2 × 10⁻¹⁸ × 3²/2² = -4.95 × 10⁻¹⁸ J

2. Ionisation energy. The question expects the Bohr formula with Z = 3 for this electron, so the ionisation energy is 4.95 × 10⁻¹⁸ J. A photon with less energy than this, that is a longer wavelength, cannot ionise the atom.
3. Find the wavelength. λ = hc/E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(4.95 × 10⁻¹⁸) = (1.989 × 10⁻²⁵)/(4.95 × 10⁻¹⁸) = 4.02 × 10⁻⁸ m.
4. So. λ ≈ 4.02 × 10⁻⁸ m, so x is 4.

The common slip: 1. Taking the electron in n = 1 gives an energy of 19.8 × 10⁻¹⁸ J and λ ≈ 1.0 × 10⁻⁸ m, so 1, not 4. Lithium's ground-state outer electron is in n = 2.

The idea to keep. Longest wavelength means least energy, and that is exactly the ionisation energy. Always check which shell the outermost electron is in before you put n into the formula.

Answer from the NTA final answer key. Solution by PrepWiser.