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

Options

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

Answer

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

### Step 1: Understand the Forces Acting on the Electron
In the Bohr model, the centripetal force required to keep the electron in circular motion is provided by the electrostatic (Coulomb) force of attraction between the positively charged proton and the negatively charged electron.

### Step 2: Write the Expression for Electrostatic Force
The electrostatic force \( F \) between the electron and the proton can be expressed using Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} \]
where:
- \( e \) is the charge of the electron (and proton),
- \( r \) is the distance between the electron and proton, which in the ground state is the Bohr radius \( a_0 \).

### Step 3: Write the Expression for Centripetal Force
The centripetal force \( F_c \) required to keep the electron in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the speed of the electron,
- \( r \) is again the radius of the orbit, which is \( a_0 \).

### Step 4: Set the Forces Equal
Since the centripetal force is provided by the electrostatic force, we can set them equal to each other:
\[ \frac{1}{4 \pi \epsilon_0} \frac{e^2}{a_0^2} = \frac{m v^2}{a_0} \]

### Step 5: Rearrange the Equation
To find \( v^2 \), we can rearrange the equation:
\[ \frac{e^2}{4 \pi \epsilon_0 a_0^2} = \frac{m v^2}{a_0} \]
Multiplying both sides by \( a_0 \):
\[ \frac{e^2}{4 \pi \epsilon_0 a_0} = m v^2 \]

### Step 6: Solve for \( v^2 \)
Now, we can solve for \( v^2 \):
\[ v^2 = \frac{e^2}{4 \pi \epsilon_0 m a_0} \]

### Step 7: Take the Square Root
Taking the square root to find \( v \):
\[ v = \sqrt{\frac{e^2}{4 \pi \epsilon_0 m a_0}} \]

### Step 8: Simplify the Expression
This can also be expressed as:
\[ v = \frac{e}{\sqrt{4 \pi \epsilon_0 m a_0}} \]

### Final Result
Thus, the speed of the electron in the ground state of a hydrogen atom is:
\[ v = \frac{e}{\sqrt{4 \pi \epsilon_0 m a_0}} \]