In which of the following series does the 121 | vedclass

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(A) The wavelength of the spectral lines of the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1

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Lyman series

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(A) The wavelength of the spectral lines of the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,$n_1 = 1$. The first line of the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.Using $R \approx 1.097 	imes 10^7 \ m^{-1}$,the wavelength is $\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 1.097 	imes 10^7 	imes 0.75 \approx 8.2275 	imes 10^6 \ m^{-1}$.Thus,$\lambda \approx 1.215 	imes 10^{-7} \ m = 121.5 \ nm$.Since the wavelength $121.5 \ nm$ falls in the ultraviolet region,it belongs to the Lyman series.

In which of the following series does the $121.5 \ nm$ line of the spectrum of the hydrogen atom lie?

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