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
- A \(13.6\)
- B \(10.5\)
- C \(3.4\)
- D \(40.8\)
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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