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JEE Mains · Physics · STD 12 - 12. atoms

An electron of a hydrogen like atom, having \(Z=4\), jumps from \(4^{\text {th }}\) energy state to \(2^{\text {nd }}\) energy state, The energy released in this process, will be \(.........eV\) \((\) Given \(Rch =13.6\,eV )\) Where \(R =\) Rydberg constant \(c=\) Speed of light in vacuum \(h =\) Planck's constant

  1. A \(13.6\)
  2. B \(10.5\)
  3. C \(3.4\)
  4. D \(40.8\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(40.8\)

Step-by-step Solution

Detailed explanation

\(\Delta E =13.6 Z ^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]\,eV\) \(=13.6 \times(4)^2\left(\frac{1}{4}-\frac{1}{16}\right)\,eV\) \(=13.6[4-1]\,eV\) \(=13.6 \times 3=40.8\,eV\)
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