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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Radius of the electron's orbit,
r = 0.5 \times 10^{-10}\,\text{m} .
• Speed of the electron,
v = 2.2 \times 10^6\,\text{m/s} .
• Charge of the electron,
e = 1.6 \times 10^{-19}\,\text{Coulomb} .
• We take
\pi = \frac{22}{7} .
Step 2: Recall the Formula for Current
In Bohr’s model, the electron’s circular motion can be treated like a tiny current loop.
The current I due to the revolving electron is given by:
I = \frac{e v}{2\pi r}
Step 3: Substitute the Values
Substitute the known values into the formula:
I = \frac{\left(1.6 \times 10^{-19}\right) \times \left(2.2 \times 10^6\right)}{2 \times \left(\frac{22}{7}\right) \times \left(0.5 \times 10^{-10}\right)}.
Step 4: Simplify the Expression
First handle the denominator step by step:
• 2 \times \frac{22}{7} = \frac{44}{7} .
• Multiply by 0.5 \times 10^{-10} = 0.5 \times 10^{-10} .
Hence the denominator becomes
\frac{44}{7} \times 0.5 \times 10^{-10}.
So,
I
= \frac{\left(1.6 \times 10^{-19}\right) \times \left(2.2 \times 10^6\right)}{\frac{44}{7} \times 0.5 \times 10^{-10}}.
Step 5: Perform the Numerical Calculation
• Numerator:
1.6 \times 10^{-19} \times 2.2 \times 10^6
= 3.52 \times 10^{-13}.
• Denominator includes
\frac{44}{7} \times 0.5
= \frac{44 \times 0.5}{7}
= \frac{22}{7}.
So, denominator
= \frac{22}{7} \times 10^{-10}.
Thus,
I
= \frac{3.52 \times 10^{-13}}{\frac{22}{7} \times 10^{-10}}
= \frac{3.52 \times 10^{-13} \times 7}{22 \times 10^{-10}}.
= \frac{24.64 \times 10^{-13}}{22 \times 10^{-10}}
= \frac{24.64}{22} \times 10^{-3}
\approx 1.12 \times 10^{-3}\,\text{A}.
Step 6: Convert to Milliampere
Since 1\,\text{A} = 10^3\,\text{mA} ,
I = 1.12 \times 10^{-3}\,\text{A}
= 1.12\,\text{mA}.
We can also express this as
112 \times 10^{-2}\,\text{mA}.
Final Answer
The current associated with the revolving electron is
112 \times 10^{-2}\,\text{mA}
\,(\text{i.e. } 1.12\,\text{mA}).
