The Rydberg constant was introduced into atomic physics near the end of the last century. When Balmer ans Rydberg independently derived empirical relations which accounted for the observed features of atomic spectra. **Rydberg Formula** was much more general, applying to the spectra of many elements. Their results may be summarized by the formula.

where both n_{1} and n_{2} are integers but n_{2} is always greater than n_{1}. R is a constant and called **Rydberg constant** and the formula is commonly written as

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Solved problems based on Rydberg formula are given below. ### Solved Examples

**Question 1: **Calculate the wavelength of the second line in the Paschen series and show that this line lies in the near infrared, that is in the infrared region near the visible.

** Solution: **

In the Paschen series n_{1}=3 and n_{2}=4, 5, 6...

Thus the second line in the Paschen series is given by setting n_{1}=3 and n_{2}=5 and

Substituting the values in the above equation we get

$\bar \nu$ = 7.799 $\times$ 10^{3}cm^{-1 }

and

$\lambda$ = 1.282 $\times$ 10^{-4}cm = 1282nm

Which is in the near infrared region.

**Question 2: **Calculate the $\bar\nu$ for the transition involving n_{1}=6 to n_{2}=3 in an hydrogen atom.

** Solution: **

Since mass of nucleus is much larger than that of electron,

R = 109737.32cm^{-1}

$\bar\nu$ = (109737.32cm^{-1})($\frac{1}{3^{2}}$ - $\frac{1}{6^{2}}$)

$\bar\nu$ = 9144.78cm^{-1}

In the Paschen series n

Thus the second line in the Paschen series is given by setting n

Substituting the values in the above equation we get

$\bar \nu$ = 7.799 $\times$ 10

and

$\lambda$ = 1.282 $\times$ 10

Which is in the near infrared region.

Since mass of nucleus is much larger than that of electron,

R = 109737.32cm

$\bar\nu$ = (109737.32cm

$\bar\nu$ = 9144.78cm