What is the ratio of the potential energy of an electron in the first excited state of a He ⁺ ion to that in the second excited state of a Be ³⁺ ion?

Step-by-step solution

For a hydrogen-like ion, the potential energy of an electron in the n^ th orbit is given by U Z^2 n^2 , where Z is the atomic number. For the first excited state of He ⁺ ion: Z₁ = 2 , n₁ = 2 U₁ 2^2 2^2 = 1 For the second excited state of Be ³⁺ ion: Z₂ = 4 , n₂ = 3 U₂ 4^2 3^2 = 16 9 The ratio of their potential energies is: U₁ U₂ = 1 16/9 = 9 16 Thus, the required ratio is 9:16 . Answer: 9:16