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AP EAMCET · PHYSICS · Atomic Physics

The principle quantum number ' \(n\) ' corresponding to the exited state of \(\mathrm{He}^{+}\)ion, if on transition to the ground state two photons in succession with wavelength \(1026 \mathrm{~A}^{\circ}\) and \(304 \mathrm{~A}^{\circ}\) are emitted ( \(\mathrm{R}=1.097 \times 10^7 \mathrm{~m}^{-1}\) )

  1. A 2
  2. B 3
  3. C 6
  4. D 4
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Answer & Solution

Correct Answer

(C) 6

Step-by-step Solution

Detailed explanation

For first photon, \(\begin{aligned} & \lambda=304 Å, \mathrm{n}_1=1 \\ & \frac{1}{\lambda}=\mathrm{Rz}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\end{aligned}\)…
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