In a hydrogen-like ion, an electron makes a transition from the n^ th excited state to the 2^ nd excited state. If the total number of observable spectral lines in the emission spectrum is 15 , the value of n is

Step-by-step solution

The principal quantum number for the 2^ nd excited state is n₁ = 2 + 1 = 3 . The principal quantum number for the n^ th excited state is n₂ = n + 1 . The difference in the principal quantum numbers is n = n₂ - n₁ = (n + 1) - 3 = n - 2 . The formula for the maximum number of spectral lines is given by n( n + 1) 2 . Equating this to the given number of lines: (n - 2)(n - 1) 2 = 15 (n - 2)(n - 1) = 30 n^2 - 3n - 28 = 0 (n - 7)(n + 4) = 0 Since n must be a positive integer, n = 7 . Answer: 7