In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If `a_(0)` is the radius of the ground state orbit, m is the mass and e is the chargeon the electron and `varepsilon_(0)` is the vacuum permittivity,the speed of the electron is

Options

A. `(e)/(sqrt(epsi_(0)a_(0)m))`
B. zero
C. `(e)/(sqrt(4piepsi_(0)a_(0)m))`
D. `(sqrt(4 epsi_(0)a_(0)m))/(e)`

Answer

To find the speed of the electron in the ground state of a hydrogen atom using the Bohr model, we can follow these steps:

### Step 1: Understand the Forces Involved
In the Bohr model, the centripetal force required to keep the electron in its circular orbit is provided by the electrostatic force of attraction between the positively charged proton and the negatively charged electron.

### Step 2: Write the Expressions for Forces
1. **Electrostatic Force (Coulomb's Law)**: The force between the electron and proton can be expressed as:
\[ F_{\text{electrostatic}} = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{a_0^2} \]
where \( e \) is the charge of the electron, \( a_0 \) is the radius of the ground state orbit, and \( \varepsilon_0 \) is the vacuum permittivity.

2. **Centripetal Force**: The centripetal force required to keep the electron moving in a circular path is given by:
\[ F_{\text{centripetal}} = \frac{mv^2}{a_0} \]
where \( m \) is the mass of the electron and \( v \) is its speed.

### Step 3: Set the Forces Equal
For the electron to remain in a stable orbit, the electrostatic force must equal the centripetal force:
\[ \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{a_0^2} = \frac{mv^2}{a_0} \]

### Step 4: Rearrange the Equation
To find \( v^2 \), we can rearrange the equation:
\[ \frac{e^2}{4 \pi \varepsilon_0 a_0^2} = \frac{mv^2}{a_0} \]
Multiplying both sides by \( a_0 \):
\[ \frac{e^2}{4 \pi \varepsilon_0 a_0} = mv^2 \]

### Step 5: Solve for \( v^2 \)
Now, we can isolate \( v^2 \):
\[ v^2 = \frac{e^2}{4 \pi \varepsilon_0 m a_0} \]

### Step 6: Take the Square Root to Find \( v \)
Finally, taking the square root gives us the speed of the electron:
\[ v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 m a_0}} \]

### Conclusion
Thus, the speed of the electron in the ground state orbit of a hydrogen atom is:
\[ v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 m a_0}} \]