Atoms & Nuclei

JEE Main level, single correct, medium

40 of 239

Question 40 of 239

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 :

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Answer: (B) 2E=U2 \mathrm{E}=\mathrm{U}

F=k(Ze)(e)r2=mv2rKE=12mv2=12K(Ze)(e)rPE=−K(Ze)(e)rTE=K(Ze)(e)2r−K(Ze)(e)r=−K(Ze)(e)2rTE=PE22TE=PE\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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