MHT CET · Chemistry · Structure of Atom
The energy of an electron in the excited hydrogen atom is \(-3.4 \mathrm{eV}\). Then according to Bohr's theory, the angular momentum of the electron in that excited state is ( \(h=\) Plank's constant \()\)
- A \(\frac{2 \pi}{h}\)
- B \(\frac{\mathrm{n} h}{2 \pi}\)
- C \(\frac{h}{\pi}\)
- D \(\frac{3 h}{2 \pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{h}{\pi}\)
Step-by-step Solution
Detailed explanation
The energy in ground state \((\mathrm{n}=1)\) is \(-13.6 \mathrm{eV}\) The energy in \(\mathrm{n}=2\) state is \(\frac{-13.6}{4}=-3.4 \mathrm{eV}\) Angular momentum \(=\frac{\mathrm{n} h}{2 \pi}=\frac{2 h}{2 \pi}=\frac{h}{\pi}\)
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