JEE Mains · Physics · STD 12 - 12. atoms
A hydrogen atom in its ground state absorbs \(10.2\ eV\) of energy. The orbital angular momentum is increased by (Given Planck constant \(h = 6.6 \times {10^{ - 34}}J - sec\))
- A \(1.05 \times {10^{ - 34}} J-sec\)
- B \(3.16 \times {10^{ - 34}} J-sec\)
- C \(2.11 \times {10^{ - 34}} J-sec\)
- D \(4.22 \times {10^{ - 34}}J-sec\)
Answer & Solution
Correct Answer
(A) \(1.05 \times {10^{ - 34}} J-sec\)
Step-by-step Solution
Detailed explanation
(a) Electron after absorbing \(10.2\ eV\) energy goes to its first excited state \((n=2)\) from ground state \((n = 1)\). \(\therefore \) Increase in momentum \( = \frac{h}{{2\pi }}\)\( = \frac{{6.6 \times {{10}^{ - 34}}}}{{6.28}} = 1.05 \times {10^{ - 34}}J{\rm{ - }}s\).
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