Question

If the wavelength of the first line of the Balmer series of hydrogen is 6561 $\stackrel{\circ }{A}$ , the wavelength of the second line of the series should be

• A. $13122\stackrel{\circ }{A}$
• B. $3280\stackrel{\circ }{A}$
• C. $4860\stackrel{\circ }{A}$
• D. $2187\stackrel{\circ }{A}$

Answer 1

The correct option is C $4860\stackrel{\circ }{A}$

The wavelength of spectral line in Balmer series is given by $\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{n^2}]$
For first line of Balmer series, n = 3
$\Rightarrow \frac{1}{{\lambda }_1}=R[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5R}{36}$; For second line n = 4.
$\Rightarrow \frac{1}{{\lambda }_2}=R[\frac{1}{2^2}-\frac{1}{4^2}]=\frac{3R}{16}$
$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{20}{27}\Rightarrow {\lambda }_1=\frac{20}{27}\times 6561=4860\stackrel{\circ }{A}$