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Physics · Atomic Physics

JEE Main 2024 — 5 April, Shift 1 — Question 38

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:

  1. Option A:

    E=2U\mathrm{E}=2 \mathrm{U}

  2. Option B:

    2E=3U2 \mathrm{E}=3 \mathrm{U}

  3. Option C:

    E=UE=U

  4. Option D:

    2E=U2 \mathrm{E}=\mathrm{U}

    Correct

Answer: D

Step-by-step solution

F=k(Ze)(e)r2=mv2rF=\frac{k(Z e)(e)}{r^{2}}=\frac{\mathrm{mv}^{2}}{r}

KE=12mv2=12 K(Ze)(e)r\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \frac{\mathrm{~K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}

PE=−K(Ze)(e)r\mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} TE=K(Ze)(e)2r−K(Ze)(e)r=−K(Ze)(e)2r\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}} TE=PE2\mathrm{TE}=\frac{\mathrm{PE}}{2}

2TE=PE2 \mathrm{TE}=\mathrm{PE}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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