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 : \(\pi=3.14]\)
- A \(1010\)
- B \(6135\)
- C \(3155\)
- D \(3845\)
Answer & Solution
Correct Answer
(C) \(3155\)
Step-by-step Solution
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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