For visible line spectrum, i.e. Balmer series `n_(1) = 2` Also for minimum energy transition `n_(2) = 2`
`:. (1)/(lambda) = R_(H)((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))`
`= R_(H)[(1)/(2^(2))- (1)/(3^(2))]`
`= 1.1 xx 10^(7)[(1)/(4)-(1)/(9)] = 1.1xx 10^(7) xx (5)/(36)`
`:. lambda = 6.55 xx 10^(-7) m`
Now `E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(6.55 xx 10^(-7))`
`= 3.03 xx 10^(-19) J`
If V electron show this transition in `1 g ` atom of `h` ythen energy released `= E xx N `
`= 3.03 xx 10^(-19) xx 6.023 xx 10^(23)`
`= 18.25 xx 10^(4) J = 182.5 kJ`