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Hydrogen spectrum

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Question

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 $\overset{0}{A}$ . The wavelength of the second member of the Balmer series (in nm) is ----

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Solution

$\begin{array}{l} W a v e l e n g t h d u r i n g t r a n s i t i o n c a n b e g i v e n a s, \\ \frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ F o r f i r s t m e m b e r o f B a l m e r S e r i e s o f H y d r o g e n : \frac{1}{λ_{1}} = R {(1)}^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) \\ F o r s e c o n d m e m b e r o f B a l m e r S e r i e s o f H y d r o g e n : \frac{1}{λ_{2}} = R {(1)}^{2} (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16} \\ ∴ \frac{λ_{2}}{λ_{1}} = \frac{\frac{16}{3 R}}{\frac{36}{5 R}} = \frac{20}{27} \\ \Rightarrow λ_{2} = \frac{20}{27} \times 6561 \overset{0}{A} = 4860 \overset{0}{A} = 486 n m \end{array}$

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