The electron in the nᵗʰ orbit of Li²⁺ is excited to (n+1) orbit using the radiation of energy 1.47 × 10⁻¹⁷ J (as shown in the diagram). The value of n is ___________
Given: R_H=2.18 × 10⁻¹⁸ J

Solution

Chemistry ma’am · 4 steps

Put Z = 3 into the Bohr energy gap, divide it out, and the bracket that is left points straight to n.

1. Energy gap. For a one-electron species, the energy of level n is Eₙ = -R_H Z²/n². The absorbed photon must exactly match the gap between levels n and n+1.

   Δ E = R_H Z²(1/n²-1/(n+1)²)

2. Put in numbers. For Li²⁺, Z = 3, so R_H Z² = 9 × 2.18 × 10⁻¹⁸ = 1.962 × 10⁻¹⁷ J. Dividing gives (1.47 × 10⁻¹⁷)/(1.962 × 10⁻¹⁷) = 0.749 ≈ 3/4.
3. Find n. You need 1/n²-1/(n+1)² = 3/4. With n = 1 this is 1 - 1/4 = 3/4, which matches. Any larger n gives a much smaller value, for example n = 2 gives only 5/36.
4. So. The electron is excited from the first orbit to the second, so n = 1.

The idea to keep. For He⁺, Li²⁺ and other one-electron ions, scale the hydrogen energy by Z². Divide Δ E by R_H Z² first, then match the bracket to a neat fraction.

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