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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 \(h = 6.6 \times {10^{ - 34}}J -  sec\))

  1. A \(1.05 \times {10^{ - 34}} J-sec\)
  2. B \(3.16 \times {10^{ - 34}} J-sec\)
  3. C \(2.11 \times {10^{ - 34}} J-sec\)
  4. D \(4.22 \times {10^{ - 34}}J-sec\)
Verified Solution

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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