Question

The wavelength of the first member of the Balmer series of hydrogen spectrum is $6563\mathring{A}$ . Calculate the wavelength of second member of the Paschen series in the same spectrum

• A. $8563\mathring{A}$
• B. $10818\mathring{A}$
• C. $6409\mathring{A}$
• D. $12818\mathring{A}$

Answer 1

Answer: (D)

For first member of Balmer series
$\frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ ... (i)
For second member of Paschen series
$\frac{1}{{\lambda }_2}=R\left(\frac{1}{3^2}-\frac{1}{5^2}\right)$ ... (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{1}{2^2}-\frac{1}{3^2}}{\frac{1}{3^2}-\frac{1}{5^2}}$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{25}}$
$=\frac{5\times 225}{16\times 36}=1.953$
$\Rightarrow {\lambda }_2=1.953\times 6563$
$=12818\mathring{A}$