Question

Calculate minumim wavelength of spectral line present in balmer series of hydrogen

Answer 1

The wave length of balmer series of lines in the hyhrogen emission spectrum is given by

$\frac{1}{wave length}$ = $R_{H}$[$\frac{1}{2²}$₋$\frac{1}{n[tex]_{i}$}²[/tex]

For minimum wave length, $n_{i}$ = 3

$R_{H}$= 1·09678×10⁷ m⁻¹

Therefore,

$\frac{1}{wave length}$ = 1·09678×10⁷[$\frac{1}{2²}$₋$\frac{1}{n[tex]_{3}$}²[/tex]

$\frac{1}{wave length}$ = 1·09678×10⁷[$\frac{1}{4}$₋$\frac{1}{9}$}

$\frac{1}{wave length}$ = 1·09678×10⁷×$\frac{5}{36}$ m⁻¹

wave length                             = $\frac{1}×1·523{1·09678×10⁷}$ m

                                                 = ·65703×10⁻⁶ m

                                                 = 657×10⁻⁹ m

                                                 = 657 nm    ( 1nm = 10⁻⁹ m)