# Answer to Question #2018 in Inorganic Chemistry for vinay

Question #2018

Calculate the spacing between the second and third spectral lines in the Balmer series.

Expert's answer

Let's use the Balmer formula:

<img src="/cgi-bin/mimetex.cgi?%5Clambda%5C%20=%20B%5Cleft%28%5Cfrac%7Bm%5E2%7D%7Bm%5E2%20-%202%5E2%7D%5Cright%29" title="\lambda\ = B\left(\frac{m^2}{m^2 - 2^2}\right)">

where for the second spectral line m = 4 , for the third m = 5.

Thus

<img src="http://latex.codecogs.com/gif.latex?%5CDelta%5Clambda%5C%20=%20364.56[nm]%5Cleft%28%5Cfrac%7B25%5E2%7D%7B25%5E2%20-%202%5E2%7D-%5Cfrac%7B16%5E2%7D%7B16%5E2%20-%202%5E2%7D%5Cright%29%20=%20364.56/7%20=%2052.08%20[nm]" title="\Delta\lambda\ = 364.56[nm]\left(\frac{25^2}{25^2 - 2^2}-\frac{16^2}{16^2 - 2^2}\right) = 364.56/7 = 52.08 [nm]">

<img src="/cgi-bin/mimetex.cgi?%5Clambda%5C%20=%20B%5Cleft%28%5Cfrac%7Bm%5E2%7D%7Bm%5E2%20-%202%5E2%7D%5Cright%29" title="\lambda\ = B\left(\frac{m^2}{m^2 - 2^2}\right)">

where for the second spectral line m = 4 , for the third m = 5.

Thus

<img src="http://latex.codecogs.com/gif.latex?%5CDelta%5Clambda%5C%20=%20364.56[nm]%5Cleft%28%5Cfrac%7B25%5E2%7D%7B25%5E2%20-%202%5E2%7D-%5Cfrac%7B16%5E2%7D%7B16%5E2%20-%202%5E2%7D%5Cright%29%20=%20364.56/7%20=%2052.08%20[nm]" title="\Delta\lambda\ = 364.56[nm]\left(\frac{25^2}{25^2 - 2^2}-\frac{16^2}{16^2 - 2^2}\right) = 364.56/7 = 52.08 [nm]">

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