Physics › Structure of Atoms and Nuclei › Bohr's Atomic Model
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total…
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 :
$\mathrm{E}=\mathrm{U}$ $2 \mathrm{E}=\mathrm{U}$ $2 \mathrm{E}=3 \mathrm{U}$ $\mathrm{E}=2 \mathrm{U}$
Solution
$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}=\frac{-\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{PE}}{2} \\ & 2 \mathrm{TE}=\mathrm{PE}\end{aligned}$
Asked in: JEE Main 2024 (05 Apr Shift 1)
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