Question

Calculate the Rydberg constant if $H{e}^{+}$ ions are known to have the wavelength difference between longest wavelength lines of Balmer and Lyman series equal to $133.7nm$.

• A. $1.095\times {10}^{5}c{m}^{-1}$
• B. $9.11\times {10}^{7}c{m}^{-1}$
• C. $2.493\times {10}^{4}c{m}^{-1}$
• D. $1.095\times {10}^{8}c{m}^{-1}$

Answer 1

The correct option is A $1.095\times {10}^{5}c{m}^{-1}$

wave length of the first line of Balmer Series
$\frac{1}{{\lambda }_1}=R{z}^2(\frac{1}{2^2}-\frac{1}{3^2})\Rightarrow {\lambda }_1=\frac{9}{5R}$
wave length of the first line of Lymen Series
$\frac{1}{{\lambda }_1^{{}^{\prime }}}=R{z}^2(\frac{1}{1^2}-\frac{1}{2^2})\Rightarrow {\lambda }_1^{{}^{\prime }}=\frac{1}{3R}$
$\therefore \Delta \lambda ={\lambda }_1-{\lambda }_1^{{}^{\prime }}=\frac{22}{15R}\Rightarrow R=\frac{22}{15\Delta \lambda }=d\frac{22}{15\times 133.7\times {10}^{-9}}$
$=1.095\times {10}^7{m}^{-1}$ $1.095\times {10}^{5}c{m}^{-1}$