Atomic StructureHard
Question
The value of Rydberg constant R, if He+ ions are known to have the wavelength difference between the first (of the longest wavelength) lines of the Balmer and Lyman series equal to 132 nm, is
Options
A.2.07 × 1016 m–1
B.1.11 × 107 m−1
C.9 × 10−8 m−1
D.1.936 × 107 m−1
Solution
$\frac{1}{\lambda_{1}} = R \times 2^{2}\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) \Rightarrow \lambda_{1} = \frac{9}{5R}$
$\frac{1}{\lambda_{2}} = R \times 2^{2}\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) \Rightarrow \lambda_{2} = \frac{1}{5R}$
From question, $\lambda_{1} - \lambda_{2}$=132 nm
$\text{Or, }\frac{9}{5R} - \frac{1}{3R} = 132 \times 10^{- 9}\text{ m } \Rightarrow \text{R=1.11} \times \text{1}\text{0}^{9}\text{ }\text{m}^{- 1}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
Photon having energy equivalent to the binding energy of 4th state of He+ ion is used to eject an electron from the meta...Which one of the following statements is not true...Rutherford′s α-particle scattering experiment eventually led to thje conclusion that :...When alpha particles are sent through a thin metal foil, most of them go straight through the foil because :...Wavelength of photon having energy 1 eV would be...