The radius of the n^ th Bohr orbit of a Be ³⁺ ion is 211.6 pm . If the radius of the second Bohr orbit of He ⁺ ion is 105.8 pm , what is the principal quantum number n for the Be ³⁺ orbit?

Step-by-step solution

According to Bohr's model, the radius of the n^ th orbit is given by: r_n = a₀ n^2 Z For the second orbit of He ⁺ ( n=2, Z=2 ): r_ He ^+ = a₀ 2^2 2 = 2a₀ = 105.8 pm For the n^ th orbit of Be ³⁺ ( Z=4 ): r_ Be ³⁺ = a₀ n^2 4 = 211.6 pm Taking the ratio of the two radii: r_ Be ³⁺ r_ He ^+ = a₀ n^2 4 2a₀ = n^2 8 Substitute the given values: 211.6 105.8 = n^2 8 2 = n^2 8 n^2 = 16 n = 4 The principal quantum number is 4 . Answer: 4