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(C) The Rydberg formula for the hydrogen atom is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$Here,$n_1 = 1$ (sta

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$91$

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(C) The Rydberg formula for the hydrogen atom is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$Here,$n_1 = 1$ (stationary state) and $n_2 = \infty$ (infinity).Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1} \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight]$Since $\frac{1}{\infty} = 0$,we have: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1} 	imes 1$$\lambda = \frac{1}{1.097 	imes 10^7} \ m \approx 9.115 	imes 10^{-8} \ m$Converting to nanometers $(nm)$: $\lambda = 9.115 	imes 10^{-8} 	imes 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

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 $= 1.097 \times 10^7 \ m^{-1}$)

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