AP EAMCET · PHYSICS · Atomic Physics
The electrostatic potential energy of the electron in an orbit of hydrogen is -6.8 eV . The speed of the electron in this orbit is ( C is speed of light in vacuum)
- A \(\frac{C}{137}\)
- B \(\frac{C}{274}\)
- C \(\frac{2 C}{137}\)
- D \(\frac{3 C}{137}\)
Answer & Solution
Correct Answer
(B) \(\frac{C}{274}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}=-6.8 \mathrm{eV}\) \(\therefore\) Total energy, \(\mathrm{E}_{\mathrm{n}}=\frac{\mathrm{U}}{2}=\frac{-6.8}{2}=-3.4 \mathrm{eV}\) \(\Rightarrow \frac{-13.6}{\mathrm{n}^2}=-3.4\) \(\therefore \quad \mathrm{n}=2\) \(\therefore\) Speed,…
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