Question

The wavelength of first line of Balmer series is $6563\stackrel{\circ }{A}$. The wavelength of first line of Lyman series will be :

• A. $1215.4\stackrel{\circ }{A}$
• B. $2500\stackrel{\circ }{A}$
• C. $7500\stackrel{\circ }{A}$
• D. $600\stackrel{\circ }{A}$

Answer 1

The correct option is A $1215.4\stackrel{\circ }{A}$

$\frac{{\lambda }_{Lyman}}{{\lambda }_{Balmer}}=\frac{(\frac{1}{2^2}-\frac{1}{3^2})}{(\frac{1}{1^2}-\frac{1}{2^2})}=\frac{5}{27}$
$[\therefore \frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{2}{n_i^2})]$
${\lambda }_{Lyman}=\frac{5}{2}\times {\lambda }_{Balmer}$
$=\frac{5}{27}\times 6563$
$=1215.4\stackrel{\circ }{A}$.