In the question number 59, the value of velocity of the revolving electron is ( radius r is `5.3xx10^-11m`)

Options

A. `1.2xx10^6ms^(-1)`
B. `2.2xx10^6ms^(-1)`
C. `3.2xx10^6ms^(-1)`
D. `4.2xx10^6ms^(-1)`

Explanation

To find the velocity of a revolving electron in a given orbit, we can use the formula derived from the balance of centripetal force and electrostatic force. Here’s the step-by-step solution:

### Step 1: Understand the relationship
The centripetal force required for an electron to revolve in a circular orbit is provided by the electrostatic force between the electron and the nucleus. The equation can be expressed as:

\[ \frac{m v^2}{r} = \frac{k e^2}{r^2} \]

Where:
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( r \) is the radius of the orbit,
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 2: Rearranging the equation
We can rearrange the equation to solve for \( v^2 \):

\[ v^2 = \frac{k e^2}{m r} \]

### Step 3: Substitute the known values
Now, substitute the known values into the equation:

- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( r = 5.3 \times 10^{-11} \, \text{m} \)

Substituting these values gives:

\[ v^2 = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31}) \times (5.3 \times 10^{-11})} \]

### Step 4: Calculate the numerator
Calculate \( (1.6 \times 10^{-19})^2 \):

\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]

Now calculate the numerator:

\[ 9 \times 10^9 \times 2.56 \times 10^{-38} = 23.04 \times 10^{-29} \]

### Step 5: Calculate the denominator
Now calculate the denominator:

\[ (9.1 \times 10^{-31}) \times (5.3 \times 10^{-11}) = 4.823 \times 10^{-41} \]

### Step 6: Divide the numerator by the denominator
Now, divide the results:

\[ v^2 = \frac{23.04 \times 10^{-29}}{4.823 \times 10^{-41}} = 4.78 \times 10^{12} \]

### Step 7: Take the square root
Now take the square root to find \( v \):

\[ v = \sqrt{4.78 \times 10^{12}} \approx 2.18 \times 10^6 \, \text{m/s} \]

### Step 8: Approximate the value
Rounding off gives us:

\[ v \approx 2.2 \times 10^6 \, \text{m/s} \]

### Final Answer
The velocity of the revolving electron is approximately \( 2.2 \times 10^6 \, \text{m/s} \).

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