The potential difference between cathode and anode in a cathode ray tube is V. The speed acquired by — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Principle of Conservation of Energy When a charged particle at rest is accelerated through an electrostatic field, the electrical work done on the particle is completely converted into its kinetic energy: $$\text{Gain in Kinetic Energy} = \text{Loss in Potential Energy}$$ $$\frac{1}{2} m v^2 = e V$$ where: * $e$ = charge of the electron * $V$ = potential difference between cathode and anode * $m$ = mass of the electron * $v$ = final speed of the electron --- ### Step 2 - Establish the Expression for Electron Speed Solving for $v$: $$v^2 = \frac{2 e V}{m}$$ $$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2e}{m}} \cdot \sqrt{V}$$ Since $e$ and $m$ are universal physical constants: $$v \propto \sqrt{V}$$ Thus, the speed acquired by the electrons is directly proportional to the square root of the potential difference ($\sqrt{V}$). --- ### Step 3 - Systematic Analysis of the Options * **Option (A) $V$:** Incorrect. It is the *kinetic energy* of the electron that is directly proportional to $V$, not the speed. * **Option (B) $\sqrt{V}$:** Correct. As derived, $v = \sqrt{\frac{2eV}{m}}$, so $v \propto \sqrt{V}$. * **Option (C) $V^2$:** Incorrect. Speed does not scale quadratically with potential difference. * **Option (D) $1/\sqrt{V}$:** Incorrect. This implies increasing voltage would decrease speed, which is physically wrong. $$\text{Correct Option: } \boxed{\text{B}}$$