Correct Option: A To determine the de Broglie wavelength associated with an electron accelerated through a potential difference, we follow these steps: 1. **Calculate the Kinetic Energy:** When an electron (charge $e$) is accelerated from rest through a potential difference $V$, it gains kinetic energy $K$ equal to the work done on it by the electric field. $K = e V$ 2. **Relate Kinetic Energy to Momentum:** For a particle with mass $m_e$ and momentum $p$, its non-relativistic kinetic energy is given by: $K = \frac{p^2}{2m_e}$ From this, we can express the momentum $p$ as: $p = \sqrt{2m_e K}$ 3. **Substitute Kinetic Energy into Momentum Equation:** Substitute the expression for kinetic energy ($K = eV$) into the momentum equation: $p = \sqrt{2m_e (eV)}$ $p = \sqrt{2m_e eV}$ 4. **Apply de Broglie Hypothesis:** According to the de Broglie hypothesis, the wavelength $\lambda$ associated with a particle is inversely proportional to its momentum $p$: $\lambda = \frac{h}{p}$ where $h$ is Planck's constant. 5. **Final Expression for Wavelength:** Substitute the expression for momentum $p$ into the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2m_e eV}}$ This derived expression matches option A. **Analysis of other options:** * **Option B) $\lambda = \frac{h}{m_e eV}$**: This would imply $p = m_e eV$, which is incorrect. Momentum is not directly $m_e$ times the kinetic energy. * **Option C) $\lambda = \frac{\sqrt{2m_e eV}}{h}$**: This is the reciprocal of the correct de Broglie wavelength, implying $p = \frac{1}{\lambda}h$ which is dimensionally incorrect if $p$ is taken as $\sqrt{2m_e eV}$. * **Option D) $\lambda = \frac{eV}{h}$**: This expression is dimensionally incorrect for wavelength and does not relate to momentum or mass in the context of de Broglie wavelength.