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JEE Mains · Chemistry · STD 11 - 2. structure of atom

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to \(\frac{\mathrm{h}^{2}}{\mathrm{xma}_{0}^{2}}\). The value of \(10 \mathrm{x}\) is ........ . \(\left(\mathrm{a}_{0}\right.\) is radius of Bohr's orbit) (Nearest integer) [Given : \(\pi=3.14]\)

  1. A \(1010\)
  2. B \(6135\)
  3. C \(3155\)
  4. D \(3845\)
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Answer & Solution

Correct Answer

(C) \(3155\)

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Detailed explanation

\(\operatorname{mvr}=\frac{\mathrm{nh}}{2 \pi}\) \(\text { K.E. }=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \pi^{2} \mathrm{mr}^{2}} \quad=\frac{4 \mathrm{~h}^{2}}{8 \pi^{2} \mathrm{~m}\left(4 \mathrm{a}_{0}\right)^{2}}\)…
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