An electron having kinetic energy 10eV is circulating in a path of radius 0.1 m in an external magnetic field of intensity `10^(-4)`T. The speed of the electron will be

Options

A. `2 xx (10^6) ms^(-1)`
B. `4.8 xx (10^6) ms^(-1)`
C. `2.0 xx (10^(12)) ms^(-1)`
D. `4.8 xx (10^(12)) ms^(-1)`

Answer

To find the speed of the electron, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and speed
The kinetic energy (KE) of an electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its speed.

### Step 2: Rearranging the kinetic energy formula
From the kinetic energy formula, we can express the speed \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

### Step 3: Substitute the known values
The kinetic energy is given as \( 10 \, eV \). To convert electron volts to joules, we use the conversion factor \( 1 \, eV = 1.6 \times 10^{-19} \, J \):
\[ KE = 10 \, eV = 10 \times 1.6 \times 10^{-19} \, J = 1.6 \times 10^{-18} \, J \]

The mass of an electron \( m \) is approximately \( 9.11 \times 10^{-31} \, kg \).

### Step 4: Calculate the speed
Now we can substitute the values into the speed equation:
\[ v = \sqrt{\frac{2 \cdot 1.6 \times 10^{-18}}{9.11 \times 10^{-31}}} \]

Calculating the numerator:
\[ 2 \cdot 1.6 \times 10^{-18} = 3.2 \times 10^{-18} \]

Now, calculating the speed:
\[ v = \sqrt{\frac{3.2 \times 10^{-18}}{9.11 \times 10^{-31}}} \]

Calculating the division:
\[ \frac{3.2 \times 10^{-18}}{9.11 \times 10^{-31}} \approx 3.51 \times 10^{12} \]

Taking the square root:
\[ v \approx \sqrt{3.51 \times 10^{12}} \approx 1.87 \times 10^6 \, m/s \]

### Step 5: Final answer
Thus, the speed of the electron is approximately:
\[ v \approx 1.87 \times 10^6 \, m/s \]