One of the emission spectral lines for be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state with n 5 5. what is the principal quantum number of the lower-energy state corresponding to this emission? (hint: the bohr model can be applied to one-electron ions. don’t forget the z factor: z 5 nuclear charge 5 atomic number.)

Explanation:

The given data is as follows.

    = 253.4 nm =      (as 1 nm = )

            = 5,         = ?

Relation between energy and wavelength is as follows.

                    E =

                       =

                       = J

                       =

Hence, energy released is .

Also, we known that change in energy will be as follows.

where, Z = atomic number of the given element

      0.02 + 0.04 =

                      =

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.