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

If shortest wavelength of hydrogen atom in Lyman series is \(x\), then longest wavelength in Balmer series of \(\text{He}^+\) is:

  1. A \(\dfrac{9x}{5}\)
  2. B \(\dfrac{36x}{5}\)
  3. C \(\dfrac{x}{4}\)
  4. D \(\dfrac{5x}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\dfrac{9x}{5}\)

Step-by-step Solution

Detailed explanation

For the shortest wavelength in the Lyman series of the hydrogen atom (\(Z=1\)), the transition is from \(n_2 = \infty\) to \(n_1 = 1\). \(\dfrac{1}{x} = R(1)^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right) = R\) This gives \(R = \dfrac{1}{x}\). For the longest wavelength in…
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