If the series limit of wavelength of the Lyma | vedclass

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(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the series limit of the Lyman series,

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$912 	imes 4 \ \mathring{A}$

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(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the series limit of the Lyman series,$n_1 = 1$ and $n_2 = \infty$.Thus,$\frac{1}{\lambda_L} = R(1 - 0) = R$,which implies $\lambda_L = \frac{1}{R} = 912 \ \mathring{A}$.For the series limit of the Balmer series,$n_1 = 2$ and $n_2 = \infty$.Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 ight) = \frac{R}{4}$.This gives $\lambda_B = \frac{4}{R} = 4 \lambda_L$.Substituting the value of $\lambda_L$,we get $\lambda_B = 4 	imes 912 \ \mathring{A}$.

If the series limit of wavelength of the Lyman series for the hydrogen atom is $912 \ \mathring{A}$,then the series limit of wavelength for the Balmer series of radiation which may be emitted is :

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