An electron in the ground state of a He ⁺ ion absorbs 8.175 10⁻¹⁸ J of energy and transitions to a higher orbit. The maximum number of spectral lines emitted when this electron returns to the ground state is ________. Given : R_H = 2.18 10⁻¹⁸ J

Step-by-step solution

First, we determine the principal quantum number n of the excited state. The energy absorbed during the transition from the ground state ( n₁ = 1 ) to an excited state ( n₂ = n ) is: E = R_H Z^2 [ 1 1^2 - 1 n^2 ] For He ⁺ , the atomic number Z = 2 . Substituting the given values: 8.175 10⁻¹⁸ = 2.18 10⁻¹⁸ (2)^2 [ 1 - 1 n^2 ] 8.175 10⁻¹⁸ = 8.72 10⁻¹⁸ [ 1 - 1 n^2 ] 1 - 1 n^2 = 8.175 8.72 1 - 1 n^2 = 0.9375 1 - 1 n^2 = 15 16 1 n^2 = 1 16 n^2 = 16 n = 4 The electron is excited to the 4th orbit ( n = 4 ). The maximum num