The frequency of radiations emitted when electron falls from `n = 4` to `n = 1` in `H-"atom"` would be (Given `E_1` for `H = 2.18 xx 10^-18 J "atom"^-1` and `h = 6.625 xx 10^-34 Js`.)

Options

A. `1.54 xx10^(15) s^(-1)`
B. `1.03xx10^(15) s^(-1)`
C. `3.08xx10^(15) s^(-1)`
D. `2.00xx10^(15)s^(-1)`

Answer

C

Explanation

(c) Ionisation energy of H `=2.18xx10^(-18) J "atom"^(-1)`
`:. E_(1)`(Energy of Ist orbit of H-atom) =`- 2.18 xx 10^(-18) J "atom"^(-1)`
`:. E_(n)=(-2.18xx10^(-18))/(n^(2)) J "atom"^(-1)`
Z=1 for H-atom
`DeltaE=E_(4)-E_(1)`
`=(-2.18xx10^(-18))/(4^(2))-(-2.18xx10^(-18))/(1^(2))`
`=-2.18xx10^(-18)xx[(1)/(4^(2))-(1)/(1^(2))]`
`DeltaE=-2.18xx10^(-18)xx-(15)/(16)`
`=+2.0437xx10^(-18) J "atom"^(-1)`
`:. v=(DeltaE)/(h)`
`=(2.0437xx10^(-18)J "atom"^(-1))/(6.625xx10^(-34) Js)`
`=3.084xx10^(15)s^(-1) "atom"^(-1)`