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(A) The frequency $
u$ of radiation emitted is given by the Rydberg formula: $
u = R_H c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Here,$R_H

Vedclass · Accepted Answer

$3.08 	imes 10^{15} \, s^{-1}$

Vedclass · Answer

(A) The frequency $
u$ of radiation emitted is given by the Rydberg formula: $
u = R_H c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Here,$R_H = 1.097 	imes 10^7 \, m^{-1}$,$c = 3 	imes 10^8 \, m/s$,$n_1 = 1$,and $n_2 = 4$.Substituting the values: $
u = (1.097 	imes 10^7) 	imes (3 	imes 10^8) 	imes \left( \frac{1}{1^2} - \frac{1}{4^2} ight)$.$
u = 3.291 	imes 10^{15} 	imes \left( 1 - \frac{1}{16} ight) = 3.291 	imes 10^{15} 	imes \frac{15}{16}$.$
u \approx 3.08 	imes 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 $h = 6.625 \times 10^{-34} \, J s$):-

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