The frequency of radiation emiited when the electron falls 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)`

Options

A. `1.54 xx 10^(15) s^(-1)`
B. `1.03 xx 10^(15) s^(-1)`
C. `3.08 xx 10^(15) s^(-1)`
D. `2.0 xx 10^(15) s^(-1)`

Answer

C

Explanation

`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)`