Centripetal acceleration of electron in the first Bohr orbit will be

Options

A. `9xx10^(22) m//s^(2)`
B. `4xx10^(22) m//s^(2)`
C. `6xx10^(22) m//s^(2)`
D. `2xx10^(22) m//s^(2)`

Explanation

To find the centripetal acceleration of an electron in the first Bohr orbit, we can use the formula for centripetal acceleration, which is given by:

\[ a_c = \frac{v^2}{r} \]

Where:
- \( a_c \) is the centripetal acceleration,
- \( v \) is the velocity of the electron,
- \( r \) is the radius of the first Bohr orbit.

### Step 1: Determine the radius of the first Bohr orbit
The radius of the first Bohr orbit (n=1) is given by the formula:

\[ r = \frac{n^2 h^2}{4 \pi^2 k m_e} \]

Where:
- \( n \) is the principal quantum number (for the first orbit, \( n=1 \)),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( k \) is Coulomb's constant (\( \frac{1}{4 \pi \epsilon_0} \)),
- \( m_e \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)).

For the first Bohr orbit, we can simplify the radius to:

\[ r = 5.29 \times 10^{-11} \, \text{m} \]

### Step 2: Calculate the velocity of the electron in the first Bohr orbit
The velocity of the electron in the first Bohr orbit can be calculated using the formula:

\[ v = \frac{e^2}{4 \pi \epsilon_0 h} \]

Where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)).

Substituting the values, we find:

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

### Step 3: Calculate the centripetal acceleration
Now we can substitute the values of \( v \) and \( r \) into the centripetal acceleration formula:

\[ a_c = \frac{(2.18 \times 10^6)^2}{5.29 \times 10^{-11}} \]

Calculating this gives:

\[ a_c \approx 9 \times 10^{22} \, \text{m/s}^2 \]

### Conclusion
Thus, the centripetal acceleration of the electron in the first Bohr orbit is approximately:

\[ a_c \approx 9 \times 10^{22} \, \text{m/s}^2 \]

### Final Answer
The correct option for this question will be option A: \( 9 \times 10^{22} \, \text{m/s}^2 \).

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