MHT CET · Physics · Atomic Physics
The potential energy of the orbital electron in the ground state of hydrogen atoms is \(-\mathrm{E}\). What is the kinetic energy?
- A \(4 \mathrm{E}\)
- B \(\frac{E}{4}\)
- C \(\frac{E}{2}\)
- D \(2 E\)
Answer & Solution
Correct Answer
(C) \(\frac{E}{2}\)
Step-by-step Solution
Detailed explanation
The potential energy of
\(\mathrm{U}(\mathrm{n}=1)=-\mathrm{E}\)
Goomd state: total energy is given by
\(\mathrm{TE}=\mathrm{U}+\mathrm{K}\)
\(\mathrm{U}=-\frac{\left(\mathrm{ze}^2\right)}{4 \pi \varepsilon_0}\)
For orbit consider force balance

\(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)
\(\therefore \mathrm{K}=\frac{\mathrm{mv}^2}{2}=\frac{1}{2}\left(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}\right)\)
Thus, \(\mathrm{K}=\frac{\mathrm{E}}{2}=\frac{\mathrm{U}}{2}\)
\(\mathrm{U}(\mathrm{n}=1)=-\mathrm{E}\)
Goomd state: total energy is given by
\(\mathrm{TE}=\mathrm{U}+\mathrm{K}\)
\(\mathrm{U}=-\frac{\left(\mathrm{ze}^2\right)}{4 \pi \varepsilon_0}\)
For orbit consider force balance

\(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)
\(\therefore \mathrm{K}=\frac{\mathrm{mv}^2}{2}=\frac{1}{2}\left(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}\right)\)
Thus, \(\mathrm{K}=\frac{\mathrm{E}}{2}=\frac{\mathrm{U}}{2}\)
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