Question

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?

Answer 1

Answer:

4

Explanation:

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

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

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