The electron in ` Li^(2+)` ions are exctied form gound stte by absorbing ` 8. 4375Rh` energy /electron . How much emission lines are expected during de-excitation of electrons to ground state ?

Answer

6

Explanation

`:. Delta = 8. 4375Rh, E_(1Li^(2+)) =- Rh xx 9`
and ` E_(4Li)^(2+) = ( -Rh xx 9)/(16) `
` :. E_1 - E_4 = ( 9Rh)/1 - ( 9Rh)/(16) = ( 135Rh)/(16) = 8. 42375 Rh` ltbRgt Thus , de-excitaition will lead form ` 4^(th)` to 1st orbit .
No. of spectral lines ltbRgt ` = sum Delta n =sum ( 4 -1) = sum 3 = 1+2+3 =6`.