if the wavelenght of thefirst line of the balmer series is 6500A.then calculate the wavelenght of second line of the balmer series .i have already asked this question .but my doubt has not been clarified .could u please explain in clear steps ...

Given : Wavelength of first line in Balmer series = 6500 A⁰
To find: Wavelength of second line in Balmer series
Solve: In Balmer series emission take place from n₂ = 3,4,5... to n₁ = 2.
As we know $\overline{\nu } = \frac{1}{\lambda }=109677 (\frac{1}{{n_1}^2}-\frac{1}{n_2^2}) c{m}^{-1}$
where $\overline{\nu } and \lambda$ are wave number and wavelength respectively. 
For first Balmer series we have n₁ = 2 and n₂ = 3.
For second Balmer series we have n₁ = 2 and n₂ = 4.
Putting n₁ = 2 and n₂ = 4 in the equation
$\frac{1}{\lambda }=109677(\frac{1}{2^2}-\frac{1}{4^2})$
$\frac{1}{\lambda }=20564.44 c{m}^{-1}$
$$\begin{gathered} {\text{λ = 0.00004862 cm}}^{-1} \\ \lambda = 4862\times {10}^{-8}c{m}^{-1} =4862 A^0 \end{gathered}$$

 

Therefore, the wavelength of second line of the Balmer series will be 4862 A⁰
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