The radius of fifth orbit of the Li⁺⁺ is __________ × 10⁻¹² m.
Take: radius of hydrogen atom = 0.51 Aᵒ

Solution

Physics sir · 4 steps

Bohr radius grows as n² and shrinks as Z: scale 0.51 Å by 25 and divide by 3.

1. Bohr radius formula. For a hydrogen-like ion with one electron, the orbit radius is rₙ = r₀ n²/Z, where r₀ is the radius of the hydrogen atom in its ground state.
2. Identify n and Z. Li⁺⁺ has lost two of its three electrons, so it is hydrogen-like with Z = 3. The fifth orbit means n = 5.

   r₅ = 0.51 Å × 5²/3 = 0.51 × 25/3 Å = 4.25 Å

3. Convert units. 1 Å = 10⁻¹⁰ m = 100 × 10⁻¹² m, so 4.25 Å = 425 × 10⁻¹² m.
4. So. The radius of the fifth orbit of Li⁺⁺ is 425 × 10⁻¹² m.

The common slip: 1275. Leaving out Z = 3 gives 0.51 × 25 = 12.75 Å, which is 1275, not 425. The extra protons in lithium pull the orbit inward.

The idea to keep. For any one-electron ion, r ∝ n²/Z. Higher orbit pushes the electron out, and a bigger nuclear charge pulls it in.

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