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} $