To find the velocity acquired by the electron, we use the principle of conservation of energy. The potential energy lost by the electron as it moves through the potential difference is converted into kinetic energy. Step 1: Equate potential energy to kinetic energy. The potential energy (PE) lost by an electron with charge $e$ moving through a potential difference $V$ is $PE = eV$. The kinetic energy (KE) gained by an electron with mass $m$ and velocity $v$ is $KE = \frac{1}{2}mv^2$. According to the conservation of energy: $$eV = \frac{1}{2}mv^2$$ Step 2: Rearrange the equation to solve for velocity $v$. Multiply both sides by 2: $$2eV = mv^2$$ Divide by $m$: $$v^2 = \frac{2eV}{m}$$ Take the square root of both sides: $$v = \sqrt{\frac{2eV}{m}}$$ This can be rewritten using the given charge-to-mass ratio $\frac{e}{m}$: $$v = \sqrt{2V \left(\frac{e}{m}\right)}$$ Step 3: Substitute the given values into the equation. Given: Potential difference, $V = 3000 \text{ V}$ Ratio of charge to mass, $\frac{e}{m} = 1.8 \times 10^{11} \text{ Ckg}^{-1}$ $$v = \sqrt{2 \times 3000 \text{ V} \times 1.8 \times 10^{11} \text{ Ckg}^{-1}}$$ $$v = \sqrt{6000 \times 1.8 \times 10^{11}}$$ $$v = \sqrt{10800 \times 10^{11}}$$ $$v = \sqrt{10.8 \times 10^3 \times 10^{11}}$$ $$v = \sqrt{10.8 \times 10^{14}}$$ Step 4: Calculate the final velocity. $$v = \sqrt{10.8} \times \sqrt{10^{14}}$$ $$v \approx 3.286 \times 10^7 \text{ ms}^{-1}$$ Rounding to two significant figures, we get: $$v \approx 3.3 \times 10^7 \text{ ms}^{-1}$$ Step 5: Compare the result with the given options. A. $3.3 \times 10^7 \text{ms}^{-1}$ B. $1.0 \times 10^7 \text{ms}^{-1}$ C. $3.3 \times 10^4 \text{ms}^{-1}$ D. $2.2 \times 10^7 \text{ms}^{-1}$ The calculated velocity matches option A. The final answer is $\boxed{\text{A. } 3.3 \times 10^7 \text{ms}^{-1}}$.