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) $$${\therefore n = 3 }{\text{Now, }\frac{1}{\lambda_{\text{req}}} = 1.09 \times 10^{7} \times 2^{2}\left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) }{\Rightarrow \lambda_{\text{req}} = 471.8\text{ nm}}$$
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