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. `1.54xx10^(15)s^(-1)`
B. `1.03xx10^(15)s^(-1)`
C. `3.08xx10^(15)s^(-1)`
D. `2.0xx10^(15)s^(-1)`

Answer

C

Explanation

According to Bohr,s theory,
`DeltaE = Z^(2)E_(1) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
where `Z^(2)E_(1) =-` Ionization enegry
Since `DeltaE = hv`,
we have `v = (DeltaE)/(h) = (Z^(2)E_(1))/(h) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
`v = (2.18xx10^(-18)J)/(6.625xx10^(-34)Js) ((1)/(1^(2))-(1)/(4^(2)))`
`= 0.329 xx 10^(16)s^(-1) ((15)/(16))`
`= 3.08 xx 10^(15) s^(-1)`
Note that we use positive sign of enegry because `v` has to be positive.