The wavelength of the radiation emitted,when | vedclass

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(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_

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$91$

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(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$For a hydrogen atom,$Z = 1$. The electron falls from infinity $(n_2 = \infty)$ to the stationary state $n_1 = 1$.Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1} 	imes (1)^2 	imes (\frac{1}{1^2} - \frac{1}{\infty^2})$$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes (1 - 0) = 1.097 	imes 10^7 \ m^{-1}$$\lambda = \frac{1}{1.097 	imes 10^7} \ m \approx 9.115 	imes 10^{-8} \ m$To convert meters to nanometers $(nm)$,multiply by $10^9$: $\lambda = 9.115 	imes 10^{-8} 	imes 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

The wavelength of the radiation emitted,when in a hydrogen atom an electron falls from infinity to stationary state $n = 1$,would be ........ $nm$ (Rydberg constant $R_H = 1.097 \times 10^7 \ m^{-1}$)

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