Correct Option: (B) **Step-by-step Solution:** 1. **Recall the de Broglie wavelength formula:** The de Broglie wavelength ($\lambda$) for a particle is given by the relation: $$\lambda = \frac{h}{p}$$ where $h$ is Planck's constant and $p$ is the momentum of the particle. 2. **Relate momentum to kinetic energy:** For a particle with mass $m$ and kinetic energy $K$, the momentum $p$ is given by: $$p = \sqrt{2mK}$$ Substituting this into the de Broglie wavelength formula, we get: $$\lambda = \frac{h}{\sqrt{2mK}}$$ 3. **Relate kinetic energy to accelerating potential for a charged particle:** When a charged particle (like an electron with charge $e$) is accelerated from rest through a potential difference $V$, its kinetic energy $K$ is equal to the work done on it by the electric field, which is: $$K = eV$$ 4. **Combine the formulae for de Broglie wavelength of an accelerated electron:** Substituting $K = eV$ into the de Broglie wavelength formula: $$\lambda = \frac{h}{\sqrt{2me V}}$$ This equation shows that the de Broglie wavelength is inversely proportional to the square root of the accelerating potential difference, i.e., $\lambda \propto \frac{1}{\sqrt{V}}$. 5. **Calculate $\lambda_1$ for potential $V$:** For the first case, the potential difference is $V$, so the de Broglie wavelength is: $$\lambda_1 = \frac{h}{\sqrt{2me V}}$$ 6. **Calculate $\lambda_2$ for potential $4V$:** For the second case, the potential difference is increased to $4V$, so the new de Broglie wavelength is: $$\lambda_2 = \frac{h}{\sqrt{2me (4V)}} = \frac{h}{\sqrt{4 \cdot (2me V)}} = \frac{h}{2\sqrt{2me V}}$$ 7. **Determine the ratio $\frac{\lambda_1}{\lambda_2}$:** Now, we take the ratio of $\lambda_1$ to $\lambda_2$: $$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2me V}}}{\frac{h}{2\sqrt{2me V}}}$$ $$\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2me V}} \times \frac{2\sqrt{2me V}}{h}$$ $$\frac{\lambda_1}{\lambda_2} = 2$$ Therefore, the ratio $\frac{\lambda_1}{\lambda_2}$ is $2$.