In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius `5.3xx10^-11m`. Find the speed of the electron in the ground state. Mass of the electron `=9.1xx10^-31 kg ` and charge of the electron `= 1.6xx10^-19C`.

Explanation

To find the speed of the electron in the ground state of the 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 circular motion is provided by the Coulomb force of attraction between the positively charged proton and the negatively charged electron.

### Step 2: Write the Equation for Centripetal Force
The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the speed of the electron,
- \( r \) is the radius of the circular path.

### Step 3: Write the Equation for Coulomb Force
The Coulomb force \( F_e \) between the electron and proton is given by:
\[ F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges of the electron and proton respectively (both have a magnitude of \( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 4: Set the Forces Equal
Since the centripetal force is provided by the Coulomb force, we can set them equal:
\[ \frac{mv^2}{r} = \frac{k \cdot (1.6 \times 10^{-19})^2}{r^2} \]

### Step 5: Rearrange the Equation
Rearranging the equation to solve for \( v^2 \):
\[ mv^2 = \frac{k \cdot (1.6 \times 10^{-19})^2}{r} \]
\[ v^2 = \frac{k \cdot (1.6 \times 10^{-19})^2}{mr} \]

### Step 6: Substitute Known Values
Now we can substitute the known values:
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( r = 5.3 \times 10^{-11} \, \text{m} \)
- \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)

Calculating \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

Now substituting into the equation:
\[ v^2 = \frac{(8.99 \times 10^9) \cdot (2.56 \times 10^{-38})}{(9.1 \times 10^{-31}) \cdot (5.3 \times 10^{-11})} \]

### Step 7: Calculate the Values
Calculating the numerator:
\[ 8.99 \times 10^9 \cdot 2.56 \times 10^{-38} = 2.303744 \times 10^{-28} \]

Calculating the denominator:
\[ 9.1 \times 10^{-31} \cdot 5.3 \times 10^{-11} = 4.823 \times 10^{-41} \]

Now dividing:
\[ v^2 = \frac{2.303744 \times 10^{-28}}{4.823 \times 10^{-41}} \approx 4.78 \times 10^{12} \]

### Step 8: Take the Square Root
Finally, taking the square root to find \( v \):
\[ v \approx \sqrt{4.78 \times 10^{12}} \approx 2.19 \times 10^6 \, \text{m/s} \]

### Final Answer
The speed of the electron in the ground state of the hydrogen atom is approximately:
\[ v \approx 2.19 \times 10^6 \, \text{m/s} \]