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Question: Answered & Verified by Expert
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :
PhysicsAtomic PhysicsJEE MainJEE Main 2024 (05 Apr Shift 1)
Options:
  • A $\mathrm{E}=\mathrm{U}$
  • B $2 \mathrm{E}=\mathrm{U}$
  • C $2 \mathrm{E}=3 \mathrm{U}$
  • D $\mathrm{E}=2 \mathrm{U}$
Solution:
1624 Upvotes Verified Answer
The correct answer is: $2 \mathrm{E}=\mathrm{U}$
$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}=\frac{-\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{PE}}{2} \\ & 2 \mathrm{TE}=\mathrm{PE}\end{aligned}$

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