In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Å. If the speed of electron is `2.2 xx 16^(6)` m/s, then the current associated with the electron will be ___________ `xx10^(-2)mA`. [Take `pi` as `(22)/(7)`]

Answer

To solve the problem, we need to find the current associated with an electron revolving in a circular orbit. The current \( I \) can be calculated using the formula:

\[ I = \frac{Q}{T} \]

where \( Q \) is the charge of the electron and \( T \) is the time period of the electron's revolution.

### Step 1: Determine the charge of the electron
The charge of an electron \( Q \) is given as:

\[ Q = 1.6 \times 10^{-19} \text{ C} \]

### Step 2: Calculate the circumference of the orbit
The circumference \( C \) of the circular orbit can be calculated using the formula:

\[ C = 2\pi r \]

Given that the radius \( r = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \), we substitute this into the formula:

\[ C = 2 \times \frac{22}{7} \times (0.5 \times 10^{-10}) \text{ m} \]

Calculating this gives:

\[ C = 2 \times \frac{22}{7} \times 0.5 \times 10^{-10} = \frac{22}{7} \times 10^{-10} \text{ m} \]

### Step 3: Calculate the time period \( T \)
The time period \( T \) can be found using the formula:

\[ T = \frac{C}{v} \]

where \( v \) is the speed of the electron, given as \( v = 2.2 \times 10^6 \, \text{m/s} \). Substituting the values we have:

\[ T = \frac{\frac{22}{7} \times 10^{-10}}{2.2 \times 10^6} \]

Calculating this gives:

\[ T = \frac{22 \times 10^{-10}}{7 \times 2.2 \times 10^6} \]

### Step 4: Substitute \( T \) into the current formula
Now we can substitute \( T \) back into the current formula:

\[ I = \frac{Q}{T} = \frac{1.6 \times 10^{-19}}{T} \]

Substituting the value of \( T \) we calculated:

\[ I = \frac{1.6 \times 10^{-19}}{\frac{22 \times 10^{-10}}{7 \times 2.2 \times 10^6}} \]

### Step 5: Simplify the expression
This simplifies to:

\[ I = \frac{1.6 \times 10^{-19} \times 7 \times 2.2 \times 10^6}{22 \times 10^{-10}} \]

Calculating this gives:

\[ I = \frac{1.6 \times 7 \times 2.2}{22} \times 10^{-19 + 10 + 6} \]

### Step 6: Final calculation
Calculating the numerical part:

\[ I = \frac{1.6 \times 7 \times 2.2}{22} = 1.12 \text{ mA} \]

Since we need the answer in terms of \( \times 10^{-2} \text{ mA} \):

\[ I = 112 \times 10^{-2} \text{ mA} \]

Thus, the final answer is:

\[ \boxed{112} \]