In an excited hydrogen atom, what is the wavelength of the spectral line emitted by an electron that jumps from an 'O’ orbital to 'L orbital?  

Concept:

The Rydberg formula is the mathematical formula to determine the wavelength of light emitted by an electron moving between the energy levels of an atom. When an electron shift from an orbital from higher energy to lower energy state, a photon of light is generated. Rydberg formula is given by

• \(\bar\nu=\frac{1}{\lambda}=R_HZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\)

where \(\bar\nu\) is wave number, \(\lambda\) is wavelength, \(R_H=\text{ Rydberg constant }=1.09677\times10^7m^{-1}\)

​Explanation:

\(\frac{1}{\lambda}=R_HZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\), Z=1 for hydrogen atom.

K=1

L=2

M=3

N=4

O=5

We have to find the wavelength of an electron jumps from O orbital to L orbital, so, \(n_1=2,n_2=5\)

\(\frac{1}{\lambda}=R_HZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\)

\(\frac{1}{\lambda}=1.09677\times10^7(\frac{1}{4^2}-\frac{1}{5^2})\)

\(\frac{1}{\lambda}=1.09277\times10^7(\frac{1}{4}-\frac{1}{25})=\frac{1.09\times10^7\times21}{100}\)

\(\frac{1}{\lambda}=\frac{1.09\times10^7\times21}{100}\)

\(\lambda=\frac{100}{21\times1.0977\times10^7}=\frac{10^7}{230517}A^o=4341A^o\)

Hence, the correct answer is \(\lambda=4341A^o\)