An electron initially at rest, is accelerated through a potential difference of `200` volt, so that it acquires a velocity `8.4 xx 10^(6)m//s`. The value of `e//m` of elctron

Options

A. `2.76 xx 10^(12) C//kg`
B. ` 1.76 xx 10^(11)C//kg`
C. `0.76 xx 10^(12)C//kg`
D. None of these

Explanation

To find the value of \( \frac{e}{m} \) (the charge-to-mass ratio) of the electron, we can use the relationship between the kinetic energy gained by the electron when it is accelerated through a potential difference. Here’s the step-by-step solution:

### Step 1: Write the equation for kinetic energy
The kinetic energy (KE) gained by the electron when it is accelerated through a potential difference \( V \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its final velocity.

### Step 2: Write the equation for electric potential energy
The electric potential energy (PE) gained by the electron when it is accelerated through a potential difference \( V \) is given by:
\[ PE = QV \]
For an electron, the charge \( Q \) is equal to the elementary charge \( e \), so:
\[ PE = eV \]

### Step 3: Set kinetic energy equal to electric potential energy
Since the kinetic energy gained is equal to the electric potential energy, we can set the two equations equal to each other:
\[ \frac{1}{2} mv^2 = eV \]

### Step 4: Rearranging the equation to find \( \frac{e}{m} \)
Rearranging the equation gives us:
\[ \frac{e}{m} = \frac{v^2}{2V} \]

### Step 5: Substitute the known values
We know:
- \( v = 8.4 \times 10^6 \, \text{m/s} \)
- \( V = 200 \, \text{V} \)

Substituting these values into the equation:
\[ \frac{e}{m} = \frac{(8.4 \times 10^6)^2}{2 \times 200} \]

### Step 6: Calculate \( v^2 \)
Calculating \( v^2 \):
\[ (8.4 \times 10^6)^2 = 70.56 \times 10^{12} = 7.056 \times 10^{13} \, \text{m}^2/\text{s}^2 \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ 2 \times 200 = 400 \]

### Step 8: Final calculation
Now, substituting back into the equation:
\[ \frac{e}{m} = \frac{7.056 \times 10^{13}}{400} \]
\[ \frac{e}{m} = 1.764 \times 10^{11} \, \text{C/kg} \]

### Final Answer
Thus, the value of \( \frac{e}{m} \) for the electron is:
\[ \frac{e}{m} \approx 1.76 \times 10^{11} \, \text{C/kg} \]

---