Answer:
4
Explanation:
Relationship between wavenumber and Rydberg constant (R) is as follows:
[tex]wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}[/tex]
Here, Z is atomic number.
R=109677 cm^-1
Wavenumber is related with wavelength as follows:
wavenumber = 1/wavelength
wavelength = 253.4 nm
[tex]wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}[/tex]
Z fro Be = 4
[tex]39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4[/tex]
Therefore, the principal quantum number corresponding to the given emission is 4.
