What frequency of radiation is emitted when the electron falls from n 4 to n 1 in an hydrogen atom?

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Text Solution

`1.54 xx 10^(15) s^(-1)``1.03 xx 10^(15) s^(-1)``3.08 xx 10^(15) s^(-1)``2.0 xx 10^(15) s^(-1)`

Answer : C

Solution : `IE. = E_(oo) - E_(1) = 0 - E_(1) = 2.18 xx 10^(-18) J "atom"^(-1)` <br> Thus, `E_(n) = - (2.18 xx 10^(-18))/(n^(2)) J "atom"^(-1)` <br> `Delta E = E_(4) - E__(1) = -2.18 xx 10^(-18) ((1)/(4^(2)) - (1)/(1^(2)))` <br> `= 2.044 xx 10^(-18) J "atom"^(-1)` <br> `Delta E = hv` <br> or `v = (Delta E)/(h) = (2.044 xx 10^(-18)j)/(6.625 xx 10^(-34)Js) = 3.085 xx 10^(15) s^(-1)`

The frequency of radiation emitted when the electron falls from n =4 to n =1 in a hydrogen atom will be given ionization energy of H =2.18 × 10 18 atom 1 and .h =6.606 × 10 34 j s A. 1.54 × 1015 s 1B. 1.03 × 1015 s 1C. 3.08 × 1015 s 1D. 2 × 1015 s 1

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