JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2024
- Hard
- MCQ
An electron rotates in a circle around a nucleus having positive charge \(\mathrm{Ze}\). Correct relation between total energy ( \(E\) ) of electron to its potential energy \((U)\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2 E=U\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\frac{\mathrm{k}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)
\(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}\)
\(\mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}\)
\(\mathrm{TE}=\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}=\frac{-\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}\)
\(\mathrm{TE}=\frac{\mathrm{PE}}{2}\)
\(\text { 2TE }=\mathrm{PE}\)
\(\text { Option (4) }\)
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