Question
The wavelength of $${H_\alpha }$$ line of Balmer series is $$X\,\mathop {\,{\text{A}}}\limits^{\text{o}} .$$ What is the $$X$$ of $${H_\beta }$$ line of Balmer series.
A.
$$X\frac{{108}}{{80}}\mathop {\,{\text{A}}}\limits^{\text{o}} $$
B.
$$X\frac{{80}}{{108}}\mathop {\,{\text{A}}}\limits^{\text{o}} $$
C.
$$\frac{1}{X}\frac{{80}}{{108}}\mathop {\,{\text{A}}}\limits^{\text{o}} $$
D.
$$\frac{1}{X}\frac{{108}}{{80}}\mathop {\,{\text{A}}}\limits^{\text{o}} $$
Answer :
$$X\frac{{80}}{{108}}\mathop {\,{\text{A}}}\limits^{\text{o}} $$
Solution :
$${H_\alpha }$$ line of Balmer series means first line of Balmer series.
$$\eqalign{
& {n_1} = 2,\,{n_2} = 3 \cr
& \bar v = \frac{1}{{{\lambda _\alpha }}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}} \cr
& \therefore \,\,{\lambda _\alpha } = \frac{{36}}{{5R}} = X \cr} $$
$${H_\beta }$$ line of Balmer series means, second line of Balmer series, $${n_1} = 2,\,{n_2} = 4.$$
$$\eqalign{
& \bar v = \frac{1}{{{\lambda _\beta }}} \cr
& = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right) \cr
& = \frac{{3R}}{{16}} \cr
& \therefore {\lambda _\beta } = \frac{{16}}{{3R}} = X \cr
& {\text{when}}\,\,\frac{{36}}{{3R}} = X \cr
& {\text{Then}}\,\,\frac{{16}}{{3R}} \cr
& = \frac{{X \times 5R \times 16}}{{36 \times 3R}} \cr
& = \frac{{80X}}{{108}}\mathop {\text{A}}\limits^{\text{o}} \cr} $$