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  • Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

2.17    Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

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Balmer formula: 

\vartheta =\frac{1}{\lambda} = R_{H}\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]

As we can note from the formula that the wavenumber is inversely proportional to the wavelength.

Hence, for the longest wavelength transition in the Balmer series of atomic hydrogen wavenumber has to be the smallest or n_{2} should be minimum i.e., n_{2} = 3.

For the Balmer series, n_{1} =2

Thus, the expression of wavenumber is given by,

\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{1}{2^2}-\frac{1}{3^2} \right ]

\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{1}{4}-\frac{1}{9} \right ]

\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{9-4}{36} \right ]

\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{5}{36} \right ] = 1.5236\times10^6\ m^{-1}

Posted by

Divya Prakash Singh

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