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 xx 1015 s^-1`
B. `1.03 xx 1015 s^-1`
C. `3.08 xx 1015 s^-1`
D. `2.0 xx 1015 s^-1`

Answer

C

Explanation

( c) Ionisation energy of `H = 2.18 xx 10^-18 J "atom"^-1`
:. `E_1` (energy of `Ist` orbit of `H-"atom"`) =` -2.18 xx 10^-18 J "atom"^-1`
:. `E_n = (-2.18 xx 10^-18)/(n^2) J atm^-1`
`Z = 1` for `H - "atom"`
`Delta E = E_n - E_1 = (-2.18 xx 10^-18)/(4^2) - (-2.18 xx 10^-18)/(1^2)`
`Delta E = -2.18 xx 10^-18 xx (-15)/(16)`
=`+ 2.0437 xx 10^-18 J at om^-1`
`v = (Delta E)/(h) = (2.04 xx 10^-18)/(6.626 xx 10^-34)`
=`3.084 xx 10^15 s^-1 at om^-1`.