An electron in a hydrogen atom is excited to a higher energy state. As it de-excites to the ground state, it emits exactly 6 distinct spectral lines. If the longest wavelength among all these emitted lines is , the shortest wavelength emitted during this de-excitation process is

Step-by-step solution

The number of distinct spectral lines emitted during de-excitation from the n^ th state to the ground state is given by n(n-1) 2 . Given that 6 lines are emitted: n(n-1) 2 = 6 n(n-1) = 12 n = 4 The possible transitions are from states up to n=4 . The longest wavelength corresponds to the minimum energy transition, which is from n=4 to n=3 . 1 = R_H ( 1 3^2 - 1 4^2 ) = R_H ( 1 9 - 1 16 ) = 7 R_H 144 The shortest wavelength corresponds to the maximum energy transition, which is from n=4 to n=1 . Let this be ' . 1 ' =