The radius of the orbital of electron in the hydrogen atom 0.5 Å. The speed of the electron is `2 xx 10^(6) m//s`. Then the current in the loop due to the motion of the electron is

Options

A. 1 mA
B. 1.5 mA
C. 2.5 mA
D. `1.5 xx 10^(-2) mA`

Answer

To find the current in the loop due to the motion of the electron in a hydrogen atom, we can follow these steps:

### Step 1: Identify the given values
- Radius of the electron's orbit (r) = 0.5 Å = \(0.5 \times 10^{-10}\) m
- Speed of the electron (v) = \(2 \times 10^6\) m/s
- Charge of the electron (Q) = \(1.6 \times 10^{-19}\) C

### Step 2: Calculate the circumference of the orbit
The circumference (C) of the circular path of the electron can be calculated using the formula:
\[ C = 2\pi r \]
Substituting the value of r:
\[ C = 2\pi (0.5 \times 10^{-10}) = \pi \times 10^{-10} \text{ m} \]

### Step 3: Calculate the time period (T) for one complete revolution
The time period (T) can be calculated using the formula:
\[ T = \frac{C}{v} \]
Substituting the values:
\[ T = \frac{\pi \times 10^{-10}}{2 \times 10^6} \]

### Step 4: Calculate the current (I)
Current (I) can be calculated using the formula:
\[ I = \frac{Q}{T} \]
Substituting the values:
\[ I = \frac{1.6 \times 10^{-19}}{T} \]
Now substituting T from Step 3:
\[ I = \frac{1.6 \times 10^{-19}}{\frac{\pi \times 10^{-10}}{2 \times 10^6}} \]
This simplifies to:
\[ I = \frac{1.6 \times 10^{-19} \times 2 \times 10^6}{\pi \times 10^{-10}} \]
\[ I = \frac{3.2 \times 10^{-13}}{\pi \times 10^{-10}} \]
\[ I \approx \frac{3.2 \times 10^{-13}}{3.14 \times 10^{-10}} \approx 1.02 \times 10^{-3} \text{ A} \]
Thus, converting to milliampere:
\[ I \approx 1.02 \text{ mA} \]

### Final Answer
The current in the loop due to the motion of the electron is approximately **1.02 mA**.

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