JEE Main 2024 5 April Morning Shift
Exam JEE Main
Year 2024
Date 5 April
Shift Morning Shift
Subject Physics
Chapter Atomic Physics JEE Main 2024 · 5 April · Morning Shift · Physics · Atomic Physics
An electron rotates in a circle around a nucleus having positive...
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 :
Options A E = U \mathrm{E}=\mathrm{U} E = U B 2 E = U 2 \mathrm{E}=\mathrm{U} 2 E = U C 2 E = 3 U 2 \mathrm{E}=3 \mathrm{U} 2 E = 3 U D E = 2 U \mathrm{E}=2 \mathrm{U} E = 2 U Correct answer B. 2 E = U 2 \mathrm{E}=\mathrm{U} 2 E = U
Step-by-step solution
F = k ( Z e ) ( e ) r 2 = m v 2 r K E = 1 2 m v 2 = 1 2 K ( Z e ) ( e ) r P E = − K ( Z e ) ( e ) r T E = K ( Z e ) ( e ) 2 r − K ( Z e ) ( e ) r = − K ( Z e ) ( e ) 2 r T E = P E 2 2 T E = P E \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} F = r 2 k ( Ze ) ( e ) = r mv 2 KE = 2 1 mv 2 = 2 1 r K ( Ze ) ( e ) PE = − r K ( Ze ) ( e ) TE = 2 r K ( Ze ) ( e ) − r K ( Ze ) ( e ) = 2 r − K ( Ze ) ( e ) TE = 2 PE 2 TE = PE
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