JEE Mains · Physics · STD 12 - 12. atoms
A light of energy \(12.75 \;eV\) is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is \(\frac{x}{\pi} \times 10^{-17} \;eVs\). The value of \(x\) is \(........\) (use \(h=4.14 \times 10^{-15} \;eVs , c=3 \times 10^8\;\left.ms ^{-1}\right)\)
- A \(800\)
- B \(828\)
- C \(830\)
- D \(895\)
Answer & Solution
Correct Answer
(B) \(828\)
Step-by-step Solution
Detailed explanation
In the ground state energy \(=-13.6\,eV\) So energy \(\quad \frac{-13.6 eV }{n^2}=-13.6+12.75\) \(\frac{-13.6 eV }{n^2}=-0.85\) \(\quad n=\sqrt{16}\) \(\qquad n=4\) \(\text { Angular momentum }=\frac{n h}{2 \pi}=\frac{4 h}{2 \pi}=\frac{2 h}{\pi}\)…
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