ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(D) \(2 E=U\)

Step-by-step Solution

Detailed explanation

\(\mathrm{F}=\frac{\mathrm{k}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)

\(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}\)

\(\mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}\)

\(\mathrm{TE}=\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}=\frac{-\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}\)

\(\mathrm{TE}=\frac{\mathrm{PE}}{2}\)

\(\text { 2TE }=\mathrm{PE}\)

\(\text { Option (4) }\)

Same subject
Explore more questions on app