When is `k_(alpha)` line in the emission spectrum of hydrogen atom obtained ? Calculate the frequency of the photon emitted during this transition.

Explanation

The `k_(alpha)` line in the emission spectrum of the hydrogen atom is obtained when the electron transition takes place between n=2 to n=1 .
We know that the energy in n-th level is
`E_n=(-13.6)/(n^2)`
`therefore E_2-E_1=(13.6)/(2^2)-(-13.6)/(1^1)=(-13.6)/(2^2)+13.6=10.2 eV`
`=10.2xx1.6xx10^(-19)J=16.32xx10^(-19)J`
Now, this difference in energy gives the energy of the photon emitted
`therefore E_2-E_1=hnu`
`therefore nu=(E_2-E_1)/(h)=(16.32xx10^(-19))/(6.63xx10^(-34))`
`therefore nu=2.46xx10^(15)Hz`