Bohr Radius

Question

The radius of the first Bohr orbit for hydrogen is 0.529 Å. What would be the radius of the second orbit for Li^{2+} ion (Z=3)?

RankGuru · Accepted Answer

The radius of the nth Bohr orbit for a hydrogen-like ion is given by r_n = a_0 × n^2 / Z, where a_0 = 0.529 Å is the first Bohr radius for hydrogen, n is the principal quantum number, and Z is the atomic number of the ion. This formula captures two competing effects: higher n increases the orbital radius while higher Z contracts it due to stronger nuclear attraction. For Li^{2+} with Z=3 and n=2: r_2 = 0.529 × (2)^2 / 3 = 0.529 × 4 / 3 = 2.116 / 3 = 0.705 Å ≈ 0.706 Å. Option 0.353 Å would correspond to r_1 of Li^{2+} (n=1): 0.529 × 1/3 = 0.176 Å — actually this value (0.353 Å) arises if n=1 and Z=1.5, which is not physical. Option 0.529 Å is the first Bohr radius for hydrogen (n=1, Z=1) and does not apply here. Option 1.058 Å would arise from using Z=1 (hydrogen) with n=2: 0.529 × 4/1 = 2.116 Å — or it could represent a factor-of-2 error. This problem is a direct application of Bohr's orbit radius formula extended to hydrogen-like ions, a core NCERT and JEE topic. Plausibility check: for Li^{2+} (Z=3) at n=2, the radius is 4/3 × 0.529 ≈ 0.706 Å, which is smaller than the hydrogen n=2 radius (4 × 0.529 = 2.116 Å) due to the higher nuclear charge, confirming the answer is physically reasonable.

Bohr Radius

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