Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Atoms

Question:

According to Bohr's theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

Options:

$+\frac{e^2}{8 \pi \varepsilon_0 r}$ and $-\frac{e^2}{4 \pi \varepsilon_0 r}$

$+\frac{8 \pi \varepsilon_0 e^2}{r}$ and $-\frac{4 \pi \varepsilon_0 e^2}{r}$

$-\frac{e^2}{8 \pi \varepsilon_0 r}$ and $-\frac{e^2}{4 \pi \varepsilon_0 r}$

$+\frac{e^2}{8 \pi \varepsilon_0 r}$ and $+\frac{e^2}{4 \pi \varepsilon_0 r}$

Correct Answer:

$+\frac{e^2}{8 \pi \varepsilon_0 r}$ and $-\frac{e^2}{4 \pi \varepsilon_0 r}$

Explanation:

P.E. = $-\frac{k e^2}{r}=-\frac{e^2}{4 \pi \varepsilon_0 r}$ ;

K.E. = $-\frac{1}{2} (P.E.) = \frac{e^2}{8 \pi \varepsilon_0 r}$