Question

The wavelength of the first member of Balmer series is $6563\stackrel{\circ }{A}$. Calculate the wavelength of second member of Lyman series.

• A. $1025.5\stackrel{\circ }{A}$
• B. $2025\stackrel{\circ }{A}$
• C. $6563\stackrel{\circ }{A}$
• D. None of these

Answer 1

The correct option is A $1025.5\stackrel{\circ }{A}$

The first orbital of Balmer series corresponds to the transition from 3 to 2 and the second member of Lyman series corresponds to the transition from 3 to 1.

Let us write the expression for the wavelength for the first member of the Balmer series.

$\frac{1}{{\lambda }_1}=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})$
$=R{Z}^2(\frac{1}{4}-\frac{1}{9})=\frac{5}{36}R{Z}^2$ -------(1)

Let us write similar expression for the second member of Lyman series.

$\frac{1}{{\lambda }_2}=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R{Z}^2(\frac{1}{1^2}-\frac{1}{3^2})$
$=R{Z}^2(1-\frac{1}{9})=\frac{8}{9}R{Z}^2$ -------(2)

Given that ${\lambda }_1=6563\stackrel{\circ }{A}$

Hence dividing equation (1) by (2),
$\frac{{\lambda }_2}{6563}=\frac{\frac{5}{36}}{\frac{8}{9}}$
And this gives ${\lambda }_2=1025.5\stackrel{\circ }{A}$