Question

$H{e}^{+}$ ions are known to have the wavelength difference between first (of the longest wavelength) lines of Balmer and Lyman series equal to $133.7nm$, then calculate the Rydberg constant $R_H$.

• A. $1.095\times {10}^{5}c{m}^{-1}$
• B. $1.095\times {10}^{6}c{m}^{-1}$
• C. $1.095\times {10}^{7}c{m}^{-1}$
• D. $1.095\times {10}^{4}c{m}^{-1}$

Answer 1

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

The general formula for the wavelength of light is
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For Balmer series, the expression becomes,
$\frac{1}{\lambda }=R\times 2^2(\frac{1}{2^2}-\frac{1}{3^2})$
$\frac{1}{\lambda }=4R(0.25-0.11)=0.56R$
$\lambda =\frac{1.8}{R}$......(1)
For laymen series
$\frac{1}{\lambda }=R\times 2^2(\frac{1}{1^2}-\frac{1}{2^2})$
$\frac{1}{\lambda }=4R(1-0.25)=3R$
$\lambda =\frac{0.33}{R}$......(2)
(1) - (2) gives
$\frac{1.8}{R}-\frac{0.33}{R}=\frac{1.47}{R}=133.7\times {10}^{-7}$
$R=1.095\times {10}^{5}c{m}^{-1}$.