In the Balmer series spectra of hydrogen , there is a line corresponding to wavelength `4344 Å` .Calculate the number of highest energy level from which electron drops to second energy level in hydrogen spectrum . `(R xx c = 3.289 xx 10^(15))`

Explanation

The wavelength of a hydrogen line in bahmer serties is `4344 Å` so
` v= (c )/(lambda) = (3 xx 10^(8) ms^(-1))/(4344 xx 10^(-10)m) = 6.907 xx 10^(4)s^(-1)`
In the balmer series the electron drop to `n = 2 , n_(1) = 2 `Let `n_(2)` be the energy level from where the electron drop to `n = 2 `
`v = Rc[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`=- 6.907 xx 10^(14) = 3.289 xx 10^(13)[(1)/(2^(2)) - (1)/(n_(2)^(2))]`
` 6.907xx10^(14)=3.289xx10^(15)[1/2^(2)-1/n_(2)^(2)]`
`(6.907 xx 10^(14))/(3.289 xx 10^(15)) = [(1)/(4) - (1)/(n_(2)^(2))] = (n_(2)^(2) - 4)/(4n_(2)^(2))`
`= (6.907)/(32.89) = (n_(2)^(2) - 4)/(4n_(2)^(2))`
`4n_(2)^(2)(6.907) = (n_(2)^(2) - 4) 32.89`
`4n_(2)^(2) xx 6.907 = 32.89 n_(2)^(2) - 4 xx 32.89 = 5262n_(2)^(2) = 131.56 n_(2)^(2) = (131.56)/(5.622)`
`n_(2) = 5`
so the electron falls from the fifth to the second energy level