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JEE Mains · Chemistry · STD 11 - 2. structure of atom

The shortest wavelength of \(H\) atom is the Lyman series is \(\lambda_{1}\). The longest wavelength in the Balmer series of \(He ^{+}\) is :

  1. A \(\frac{5 \lambda_{1}}{9}\)
  2. B \(\frac{27 \lambda_{1}}{5}\)
  3. C \(\frac{9 \lambda_{1}}{5}\)
  4. D \(\frac{36 \lambda_{1}}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{9 \lambda_{1}}{5}\)

Step-by-step Solution

Detailed explanation

As we know \(\Delta E=\frac{h c}{\lambda}\) So \(\quad \lambda=\frac{ hc }{\Delta E } \quad\) for \(\lambda\) minimum i.e. shortest; \(\Delta E=\) maximum for Lyman series \(n=1 \&\) for \(\Delta E_{\text {max }}\) Transition must be form \(n=\infty\) to \(n=1\) So…
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