The first line of the Balmer series has a wav | vedclass

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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \r

Vedclass · Accepted Answer

$1215.4$

Vedclass · Answer

(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:$\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$.Given $\lambda_{Balmer} = 6563 \mathring{A}$,so $\frac{1}{6563} = \frac{5R}{36} \implies R = \frac{36}{5 	imes 6563}$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.Substituting the value of $R$:$\frac{1}{\lambda_{Lyman}} = \frac{3}{4} 	imes \left( \frac{36}{5 	imes 6563} ight) = \frac{27}{5 	imes 6563} = \frac{27}{32815}$.$\lambda_{Lyman} = \frac{32815}{27} \approx 1215.4 \mathring{A}$.

The first line of the Balmer series has a wavelength of $6563 \mathring{A}$. What will be the wavelength of the first member of the Lyman series in $\mathring{A}$?

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