In the Bohrs model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is:

The correct option is -1.

Let us understand why this is the correct answer step-by-step using Bohr's model of the hydrogen atom.

In Bohr's model, the electron revolves around the nucleus (proton) in a stable circular orbit of radius $r$. The electrostatic force of attraction between the positively charged nucleus and the negatively charged electron provides the necessary centripetal force for this circular motion.

Mathematically, we can write:
$\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$
where $m$ is the mass of the electron, $v$ is its orbital velocity, $e$ is the elementary charge, and ${\epsilon }_0$ is the permittivity of free space.

From this relation, we can express the quantity $m{v}^2$ as:
$m{v}^2=\frac{e^2}{4\pi {\epsilon }_{0}r}$

Now, let us write the expressions for Kinetic Energy (KE), Potential Energy (PE), and Total Energy (E):

1. Kinetic Energy (KE):
The kinetic energy of the electron is given by:
$\text{KE}=\frac{1}{2}m{v}^2$
Substituting the value of $m{v}^2$ from the force equation, we get:
$\text{KE}=\frac{e^2}{8\pi {\epsilon }_{0}r}$

2. Potential Energy (PE):
The electrostatic potential energy of the electron-nucleus system is:
$\text{PE}=-\frac{e^2}{4\pi {\epsilon }_{0}r}$

3. Total Energy (E):
The total energy of the electron in the orbit is the sum of its kinetic and potential energies:
$E=\text{KE}+\text{PE}$
Substituting the expressions for KE and PE:
$E=\frac{e^2}{8\pi {\epsilon }_{0}r}-\frac{e^2}{4\pi {\epsilon }_{0}r}$
$E=-\frac{e^2}{8\pi {\epsilon }_{0}r}$

Comparing the expressions for Kinetic Energy (KE) and Total Energy (E), we can see that:
$E=-\text{KE}$

Therefore, the ratio of the kinetic energy to the total energy of the electron is:
$\frac{\text{KE}}{E}=\frac{\text{KE}}{-\text{KE}}=-1$

This confirms that the ratio is indeed -1.