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 ${ n }_{ f } = 2 and { n }_{ i } = 3$

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

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

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

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

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

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

Answer:

Mark please please please please please please please please please please