Question

The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (atom number Z) is equal to $\Delta \lambda$. The value of the Rydberg constant for the given atom is :

• A. $\frac{5}{31}\frac{1}{\Delta \lambda Z^2}$
• B. $\frac{5}{36}\frac{Z^2}{\Delta \lambda }$
• C. $\frac{31}{5}\frac{1}{\Delta \lambda Z^2}$
• D. Given information is not sufficient

Answer 1

The correct option is C

$\frac{31}{5}\frac{1}{\Delta \lambda Z^2}$

$\Delta \lambda ={\lambda }_1-{\lambda }_2$

$=\frac{1}{R[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]z^2}-\frac{1}{R[1-0]Z^2}$

On solving, $R=\frac{31}{5}\frac{1}{\Delta \lambda .Z^2}$