If 917 mathring A is the lowest wavelength of | vedclass

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(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2}

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$3668$

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(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the lowest wavelength (series limit),$n_2 = \infty$.For the Lyman series,$n_1 = 1$. Thus,$\frac{1}{\lambda_L} = R Z^2 \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R Z^2$.Given $\lambda_L = 917 \mathring A$,we have $R Z^2 = \frac{1}{917}$.For the Balmer series,$n_1 = 2$. Thus,$\frac{1}{\lambda_B} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} ight] = \frac{R Z^2}{4}$.Substituting $R Z^2 = \frac{1}{917}$,we get $\frac{1}{\lambda_B} = \frac{1}{4 	imes 917}$.Therefore,$\lambda_B = 4 	imes 917 = 3668 \mathring A$.

If $917 \mathring A$ is the lowest wavelength of the Lyman series,then the lowest wavelength of the Balmer series will be .......$\mathring A$.

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