One of the emission spectral lines for be3+ has a wavelength of 466.4 nm for an electronic transmission that begins in the state with n = 6. what is the principal quantum number of the lower-energy state corresponding to this emission?
The answer is -6.05 x 10 ^-20. In order to get this answer, you have to use the Rydberg formula of -R(1/n ^2). Rydberg constant is -2.178x10 ^ -18 n= 6 so the solution is =-2.178 x10 ^ -18 (1/ 6^2) =-6.05 x 10 ^ -20
