Question

In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy of the electron to its potential energy, is equal to,

• A. $\frac{1}{2}$
  
• B. $2$
• C. $-\frac{1}{2}$
  
• D. $-2$

Answer 1

The correct option is C $-\frac{1}{2}$


The kinetic and potential energy of an electron in hydrogen atom, are given by,

$K.E.=\frac{m{e}^4}{8{ϵ}_0^2{h}^2{n}^2}$

$P.E.=\frac{-m{e}^4}{4{ϵ}_0^2{h}^2{n}^2}$

$\Rightarrow \frac{K.E.}{P.E.}=-\frac{1}{2}$

Hence, $\left(C\right)$ is the correct answer.
$$\begin{array}{c}\text{Why this Question ?}\\ \text{This Question is from topic Bohr's third postulates and it will clear Bohr's third postulate concept, regarding energies of an electron}\end{array}$$
Note: $|T.E|=|K.E|=|\frac{1}{2}P.E|$