In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state `n` is ……..

Explanation

To find the ratio of the kinetic energy to the total energy of the electron in a quantum state \( n \) in the Bohr model of the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kinetic Energy (KE)**:
In the Bohr model, the kinetic energy (KE) of the electron in orbit is given by the formula:
\[ KE = \frac{k e^2}{2r} \]
where \( k \) is Coulomb's constant, \( e \) is the charge of the electron, and \( r \) is the radius of the orbit.

2. **Understanding Total Energy (TE)**:
The total energy (TE) of the electron in the orbit is the sum of its kinetic energy and potential energy (PE). The potential energy for the electron in the hydrogen atom is given by:
\[ PE = -\frac{k e^2}{r} \]
Therefore, the total energy is:
\[ TE = KE + PE = \frac{k e^2}{2r} - \frac{k e^2}{r} \]
Simplifying this gives:
\[ TE = \frac{k e^2}{2r} - \frac{2k e^2}{2r} = -\frac{k e^2}{2r} \]

3. **Finding the Ratio of KE to TE**:
Now, we can find the ratio of kinetic energy to total energy:
\[ \frac{KE}{TE} = \frac{\frac{k e^2}{2r}}{-\frac{k e^2}{2r}} = -1 \]

4. **Taking the Magnitude**:
Since we are interested in the ratio in terms of magnitude, we take the absolute value:
\[ \left| \frac{KE}{TE} \right| = 1 \]

5. **Conclusion**:
Therefore, the ratio of the kinetic energy to the total energy of the electron in a quantum state \( n \) is:
\[ \frac{KE}{TE} = 1 \]

### Final Answer:
The ratio of the kinetic energy to the total energy of the electron in a quantum state \( n \) is \( 1 \).