The frequency of the radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 xx 10^(-18) J "atom"^(-1) and h = 6.625 xx 10^(-34) Js)

The frequency of the radiation emitted when the electron falls from n

1.	The frequency of the radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 xx 10^(-18) J "atom"^(-1) and h = 6.625 xx 10^(-34) Js)
Answer» `1.54 XX 10^(15) s^(-1)` `1.03 xx 10^(15) s^(-1)` `3.08 xx 10^(15) s^(-1)` `2.0 xx 10^(15) s^(-1)` SOLUTION :`IE. = E_(oo) - E_(1) = 0 - E_(1) = 2.18 xx 10^(-18) J "atom"^(-1)` Thus, `E_(n) = - (2.18 xx 10^(-18))/(n^(2)) J "atom"^(-1)` `DELTA E = E_(4) - E__(1) = -2.18 xx 10^(-18) ((1)/(4^(2)) - (1)/(1^(2)))` `= 2.044 xx 10^(-18) J "atom"^(-1)` `Delta E = hv` or `v = (Delta E)/(h) = (2.044 xx 10^(-18)j)/(6.625 xx 10^(-34)JS) = 3.085 xx 10^(15) s^(-1)`