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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 |
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Answer» Solution :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` |
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