In hydrogen atom, the electron is moving round the nucleus with velocity `2.18xx10^(6)ms^(-1)`in an orbit of radius 0.528 A. The acceleration of the electron is .

Options

A. `9 xx 10^(18) ms^(-2)`
B. `9 xx 10^(22) ms^(-2)`
C. `9 xx 10^(-22) ms^(-2)`
D. `9 xx 10^(12) ms^(-2)`

Answer

To find the acceleration of the electron in a hydrogen atom, we can use the formula for centripetal acceleration. Here's a step-by-step solution:

### Step 1: Identify the given values
- Velocity of the electron, \( v = 2.18 \times 10^6 \, \text{m/s} \)
- Radius of the orbit, \( r = 0.528 \, \text{Å} \)

### Step 2: Convert the radius from angstroms to meters
1 angstrom (Å) = \( 1 \times 10^{-10} \, \text{m} \)

So,
\[ r = 0.528 \, \text{Å} = 0.528 \times 10^{-10} \, \text{m} = 5.28 \times 10^{-11} \, \text{m} \]

### Step 3: Use the formula for centripetal acceleration
The formula for centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]

### Step 4: Substitute the values into the formula
Substituting the values of \( v \) and \( r \):
\[ a_c = \frac{(2.18 \times 10^6)^2}{5.28 \times 10^{-11}} \]

### Step 5: Calculate \( v^2 \)
Calculating \( v^2 \):
\[ (2.18 \times 10^6)^2 = 4.7524 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 6: Substitute \( v^2 \) into the acceleration formula
Now substituting \( v^2 \) back into the formula:
\[ a_c = \frac{4.7524 \times 10^{12}}{5.28 \times 10^{-11}} \]

### Step 7: Perform the division
Calculating the division:
\[ a_c = 4.7524 \times 10^{12} \div 5.28 \times 10^{-11} = 8.98 \times 10^{22} \, \text{m/s}^2 \]

### Final Result
The acceleration of the electron is approximately:
\[ a_c \approx 9 \times 10^{22} \, \text{m/s}^2 \]