Question

Calculate the wavelength of the first spectral in the Balmar senes of hydrogen spectrum.

$(R_H=109678{cm}^{-1})$

Answer 1

The 1st spectral line in Balmer series of H Spectrum is from n=3 to n=2

$\frac{1}{\lambda }$ = $R$ x $($$\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$$)$

Here $n_1$=2 and $n_2$=3 and $R$= 109678 ${cm}^{-1}$

So, $\frac{1}{\lambda }$ = 109678 x ${10}^2$ x $($$\frac{1}{2^2}$ - $\frac{1}{3^2}$$)$ $m^{-1}$

$\frac{1}{\lambda }$ = 15233 x ${10}^2$ $m^{-1}$

$\lambda$ = 656.1 nm