Question

The first member ($H_{\alpha}$ line) of the Balmer series of hydrogen has a wavelength of $6563\AA$. Calculate the wavelength of the second member ($H_{\beta}$ line).

Answer 1

"The wavelength of the spectral lines of hydrogen spectrum are given by the formula

$\frac { 1 }{ \lambda} \quad =\quad R\left( \frac { 1 }{ { n }_{ f }^{ 2 } } \quad -\quad \frac {1}{{n}_{i}^{2}} \quad\right)$

R = Rydberg constant

For Balmer series ${ n }_{ f } =2\quad and \quad { n }_{ i } = 3$

$\frac {1}{ {\lambda}_{1}} \quad =\quad R\left( \frac {1}{ 2^{2}} \quad -\quad \frac {1}{ 3^{2}} \quad\right)$

$\frac {1}{ {\lambda}_{1}} \quad =\quad \frac {5R}{36} \quad ............(i)$

For Lyman series ${n}_{f} = 1\quad and \quad {n}_{i} = 2$

$\frac {1}{ { \lambda}_{2}} \quad =\quad R\left( \frac {1 }{1^{2}} \quad -\quad \frac {1}{ 2^{2}} \quad\right)$

\$frac {1}{ {\lambda}_{2}} \quad =\quad \frac {3R}{4} \quad ............(ii)$

From equation (i) and (ii)

$\frac { {\lambda}_{1}}{ {\lambda}_{2}} \quad =\quad \frac {5R}{36} \quad \times \quad \frac {4}{3R}$

$\frac {{\lambda}_{1}}{ {\lambda}_{2}} \quad =\quad \frac {5}{27}$

From the given, ${\lambda}_{ 1 }$ = 6563

Substitute the value

${\lambda}_{ 1 }\quad =\quad \frac {5}{27} \quad \times \quad 6563\quad =\quad 1215\quad \AA$"

Answer 2

Hey mate.....☺✌☺
$\huge\blue{1215\:°A}$
is the required WAVELENGTH