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. `9xx10^(18)ms^(-2)`
B. `9xx10^(22)ms^(-2)`
C. `9xx10^(-22)ms^(-2)`
D. `9xx10^(12)ms^(-2)`

Explanation

To find the acceleration of the electron in a hydrogen atom moving in a circular orbit, we can use the formula for centripetal acceleration. Here’s how to solve the problem step by step:

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

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

### Step 3: Substitute the values into the formula
Now we can substitute the values of \( v \) and \( r \) into the formula:
\[ a = \frac{(2.18 \times 10^6)^2}{0.528 \times 10^{-10}} \]

### Step 4: Calculate \( v^2 \)
First, calculate \( v^2 \):
\[ v^2 = (2.18 \times 10^6)^2 = 4.7524 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 5: Substitute \( v^2 \) into the acceleration formula
Now substitute \( v^2 \) into the acceleration formula:
\[ a = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}} \]

### Step 6: Calculate the division
Now perform the division:
\[ a = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}} = 8.98 \times 10^{22} \, \text{m/s}^2 \]

### Step 7: Round the result
Rounding the result gives us:
\[ a \approx 9 \times 10^{22} \, \text{m/s}^2 \]

### Final Answer
The acceleration of the electron is approximately \( 9 \times 10^{22} \, \text{m/s}^2 \).

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