An electron is accelerated through a potential difference of 1000 volt. Its velocity is nearly

Options

A. `3.8xx10^7m//s`
B. `1.9 xx10^6m//s`
C. `1.9 xx10^7m//s`
D. `5.7 xx10^7m//s`

Answer

To find the velocity of an electron accelerated through a potential difference of 1000 volts, we can use the following steps:

### Step 1: Understand the relationship between potential difference and kinetic energy
The kinetic energy (KE) gained by the electron when it is accelerated through a potential difference (V) is equal to the work done on it, which can be expressed as:
\[ KE = e \cdot V \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) coulombs)
- \( V \) is the potential difference (in this case, 1000 volts)

### Step 2: Express kinetic energy in terms of mass and velocity
The kinetic energy can also be expressed in terms of the mass (m) and velocity (v) of the electron:
\[ KE = \frac{1}{2} m v^2 \]

### Step 3: Set the two expressions for kinetic energy equal to each other
Equating the two expressions for kinetic energy gives:
\[ e \cdot V = \frac{1}{2} m v^2 \]

### Step 4: Rearrange the equation to solve for velocity
Rearranging the equation to solve for velocity \( v \) gives:
\[ v^2 = \frac{2 e V}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2 e V}{m}} \]

### Step 5: Substitute known values into the equation
Now we can substitute the known values into the equation:
- \( e = 1.6 \times 10^{-19} \) C
- \( V = 1000 \) V
- \( m = 9.1 \times 10^{-31} \) kg

Substituting these values:
\[ v = \sqrt{\frac{2 \cdot (1.6 \times 10^{-19}) \cdot (1000)}{9.1 \times 10^{-31}}} \]

### Step 6: Calculate the value
Calculating the numerator:
\[ 2 \cdot (1.6 \times 10^{-19}) \cdot (1000) = 3.2 \times 10^{-16} \]

Now, calculating the entire expression:
\[ v = \sqrt{\frac{3.2 \times 10^{-16}}{9.1 \times 10^{-31}}} \]

Calculating the division:
\[ \frac{3.2 \times 10^{-16}}{9.1 \times 10^{-31}} \approx 3.51 \times 10^{14} \]

Taking the square root:
\[ v \approx \sqrt{3.51 \times 10^{14}} \approx 1.875 \times 10^{7} \text{ m/s} \]

### Step 7: Round the final answer
Rounding \( 1.875 \times 10^{7} \) to two significant figures gives approximately \( 1.9 \times 10^{7} \) m/s.

### Final Answer
The velocity of the electron is nearly \( 1.9 \times 10^{7} \) m/s.

---