JEE Mains · Chemistry · STD 11 - 2. structure of atom
If the shortest wavelength in Lyman series of hydrogen atom is \(A\), then the longest wavelength in Paschen series of \(He^+\) is
- A \(\frac{{5A}}{9}\)
- B \(\frac{{9A}}{5}\)
- C \(\frac{{36A}}{5}\)
- D \(\frac{{36A}}{7}\)
Answer & Solution
Correct Answer
(D) \(\frac{{36A}}{7}\)
Step-by-step Solution
Detailed explanation
For Lyman series (short wavelength) \({n_1} = 1,\,{n_2} = \infty \) \(\frac{1}{\lambda } = R{z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)\) \( \Rightarrow \frac{1}{A} = {1^2}\,R\left( {\frac{1}{1} - \frac{1}{\infty }} \right) \Rightarrow \frac{1}{A} = R\) Longest…
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