Question

If the wavelength of first member of Balmer series of hydrogen spectrum is $6564\dot{A}$, the wavelength of second member of Balmer series will be:

• A. $1215\dot{A}$
• B. $4862\dot{A}$
• C. $6050\dot{A}$
• D. $5892\dot{A}$

Answer 1

The correct option is B $4862\dot{A}$

$\frac{1}{{\lambda }_1}\propto [\frac{1}{2^2}-\frac{1}{3^2}]$
$\frac{1}{{\lambda }_1}\propto \frac{5}{36}$ . . . .(1)
$\frac{1}{{\lambda }_2}\propto [\frac{1}{4}-\frac{1}{16}]$
$\frac{1}{{\lambda }_2}\propto \left[\frac{3}{16}\right]$ . . . .(2)
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{5\times 16}{36\times 3}=\frac{20}{27}$
${\lambda }_2=\frac{20}{27}\times 6564=4862\dot{A}$