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A light is passed through a sample of a single elecrton excited spice which causes further excitation to some higher energy level and it emit a photon of energy 12.09 eV in back trabsition to some lower energy higher energy level .Then (i) Fing=d out the atomic number of specie if Debroglie wave lenght of electron in first excited state is 221.4 pm. (ii) Find out the value of higher energy level and lower energy level in back transition . |
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Answer» Correct Answer - (i)`Z=3`(ii)`n_(1);n_(2)=9` (i) `lambda=((150)/(V))^(1//2)Å` `lambda^(2)=(150)/(V)xx10^(-20)` `V=(150)/(221.54)xx(10^(-20))/(10^(-24))=30.6` `therefore`(T.E) First excited state=-`13.6((Z^(2))/2^(2))``=-30.6 eV` `Z=3 Ans` (ii) `E_(n2)-E_(n1)`= `DeltaE=12.09=13.6xx9((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `n_(1)=3`, `n_(2)=9` |
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