Question

The first line of the Balmer series in the hydrogen spectrum has a wavelength of $6564\mathring{A}$. Calculate the wavelength of the first line of Lyman series in the same spectrum.

• A. $1215\mathring{A}$
• B. $1222\mathring{A}$
• C. $1346\mathring{A}$
• D. $122.5\mathring{A}$

Answer 1

The correct option is A $1215\mathring{A}$

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

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

Where $R$ = Rydberg Constant

For the first line of the Balmer series, $n_f=2$ and $n_i=3$

$\frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})$

$\frac{1}{{\lambda }_1}=\frac{5R}{36}$ ................... (1)

For the first line of the Lyman series, $n_f=1$ and $n_i=2$

$\frac{1}{{\lambda }_1^{\prime }}=R(\frac{1}{1^2}-\frac{1}{2^2})$

$\frac{1}{{\lambda }_1^{\prime }}=\frac{3R}{4}$ .......................(2)

From equation (1) and (2), we get

$\frac{{\lambda }_1}{{\lambda }_1^{\prime }}=\frac{5R}{36}$ X $\frac{4}{3R}$

$\frac{{\lambda }_1}{{\lambda }_1^{\prime }}=\frac{5}{27}$

That is, ${\lambda }_1=\frac{5}{27}{\lambda }_1^{\prime }$

Given that ${\lambda }_1^{\prime }=6564\mathring{A}$

So, ${\lambda }_1=\frac{5}{27}$ X $6564$

${\lambda }_1=1215\mathring{A}$