Question

An electron at rest is accelerated through a potential difference of 200 V. If the electron acquires a velocity of \(8.4 \times 10^8\) cm/s, its \(e/m\) ratio is

• A. \(1.76 \times 10^{11}\) C/kg
• B. \(2.5 \times 10^8\) C/kg
• C. \(1.5 \times 10^8\) C/kg
• D. \(9.8 \times 10^9\) C/kg

Hint:

\(eV = \frac{1}{2}mv^2 \Rightarrow \frac{e}{m} = \frac{v^2}{2V}\). Don't forget to convert cm/s to m/s before substituting.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Energy: \(eV = \dfrac{1}{2}mv^2\), so \(\dfrac{e}{m} = \dfrac{v^2}{2V}\).
Step 2: Detailed Explanation:
\(v = 8.4 \times 10^8\) cm/s \(= 8.4 \times 10^6\) m/s
\[ \frac{e}{m} = \frac{v^2}{2V} = \frac{(8.4 \times 10^6)^2}{2 \times 200} = \frac{70.56 \times 10^{12}}{400} = 1.764 \times 10^{11} \text{ C/kg} \]
Step 3: Final Answer:
\(e/m = \mathbf{1.76 \times 10^{11}}\) C/kg.