One of the emission spectral lines for be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state with n=5. what is the principal quantum number of the lower-energy state corresponding to this emission?

4

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

Z fro Be = 4

Therefore, the principal quantum number corresponding to the given emission is 4.