AP EAMCET · PHYSICS · Atomic Physics
An electron in the hydrogen atom excites from 2 nd orbit to 4 th orbit then the change in angular momentum of the electron is
(Planck's constant \(h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}\) )
- A \(2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
- B \(1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
- C \(0.57 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
- D \(4.22 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)
Step-by-step Solution
Detailed explanation
Change in angular momentum of the electron in \(\mathrm{H}\)-atom when it is excited from 2nd orbit to 4 th orbit is given as \(\Delta L=L_2-L_1\) Where, \(L_2\) is angular momentum of 4 th excited state \((n=5)\) and \(L_1\) is angular moment of 2 nd excited ștate \((n=3)\).…
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