Find the binding energy of `L` electron in titanium if the wavelength difference between the first line of the `K` series and its short-wave cut-off is `Delta lambda= 26p m`

Explanation

By Moseley's law
` ħ omega=(2 pi ħc)/(lambda)E_(k)-E_(L)=(3)/(4) ħR(Ƶ-1)^(2)`
where `-E_(k)` is the energy of the `K` electron and `-E_(L)` of the `L` electron. Also the enrgy of the line corresponding to the short wave cut off of the `K` series is
`E_(k)=(2piħc)/(lambda-Delta lambda)=(2 piħc)/((2pic)/(omega)-Delta lambda)`
`=(ħ)/((1)/(omega)-(Delta lambda)/(2 pic))=(ħ omega)/(1-(omega Delta lambda)/(2 pic))`
Hence `E_(L)=(ħomega)/(1-(omegaDeltalambda)/(2 pi c))-ħomega=(ħomega)/((2 pic)/(omegaDeltalambda)-1)`
Substitution gives for titanium `(Ƶ=22)`
`omega= 6.85xx10^(18)s^(-1)`
and hence `E_(L)= 0.47keV`