Atomic StructureHard
Question
The wavelength of the first line of the He+ ion spectral series whose interval between the extreme lines is 2.725 × 106 m–1 is (R = 1.09 × 107 m−1)
Options
A.471.82 nm
B.4718.2 nm
C.1019.37 nm
D.165.14 nm
Solution
$\bar{\upsilon} = 2.725 \times 10^{6} = 1.09 \times 10^{7} \times 1^{2}\left( \frac{1}{(n + 1)^{2}} - \frac{1}{\infty^{2}} \right) $$${\therefore n = 3 }{\text{Now, }\frac{1}{\lambda_{\text{req}}} = 1.09 \times 10^{7} \times 2^{2}\left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) }{\Rightarrow \lambda_{\text{req}} = 471.8\text{ nm}}$$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following ...An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum required for it to es...One mole of a monoatomic real gas satisfied the equation p(V− b) = RT where b is a constant. Therelationship of in...Which of the following compounds, conjugate base is stablized by d-orbital resonance -...Hyperconjugation involves overlap of the following orbitals...