In the Balmer series spectra of hydrogen , there is a line corresponding to wavelength `4344` Å. Calculate the number of highest orbits from which electron can drop to other greater lines. `(R xx c = 3.289 xx 10^(15))`

Explanation

`v= (c )/(lambda), lambda = 4344 Å= 4344 xx 10^(-10)m`
`v = (3 xx 10^(8) m s^(-1))/(4344 xx 10^(-10)m) = 6.907 s^(-1)xx10^(14)s^(-1)`
In the balmer series s drop to `n = 2,i.e. n_(1) = 2` is the energy levels from where e' drop to `n = 2`
`v = Rc[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`6.907 xx 10^(14) = 3.289 xx 10^(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))`
`4n_(2)^(2) = 6.907 xx (n_(2)^(2) - 4) xx 3.289 xx 10`
`4n_(2)^(2) xx 6.907 = 32.89 n_(2)^(2) - 4 xx 32.89`
`5.262n_(2)^(2) = 131.56 ,n_(2)^(2) = (131.56)/(5.262) ,n_(2) = 5`
so c falls from fifth to second energy level