If {lambda _{max }} is 6563,mathring{A} for t | vedclass

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(A) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \do

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$\lambda = \frac{16}{3R}$

Vedclass · Answer

(A) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$ For the first line (maximum wavelength),$n = 3$: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$. For the second line,$n = 4$: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16}$. Therefore,$\lambda = \frac{16}{3R}$.

If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

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