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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`.)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» Correct Answer - C 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. |
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