To solve this problem, we will use the de Broglie wavelength formula and the kinetic energy of an electron.$\\$The de Broglie wavelength $\lambda$ is given by:$\\ \lambda = \frac{h}{p} \\$where $h$ is Planck's constant and $p$ is the momentum of the electron.$\\$The momentum $p$ of an electron accelerated through a potential difference $V$ is:$\\ p = \sqrt{2m_e e V} \\$where $m_e$ is the mass of the electron and $e$ is the charge of the electron.$\\$Substituting $p$ in the de Broglie wavelength formula:$\\ \lambda = \frac{h}{\sqrt{2m_e e V}} \\$Rearranging to solve for $V: \\ V = \frac{h^2}{2m_e e \lambda^2} \\$Given:$\\ \lambda = 1.227 \times 10^{-2}$ nm$= 1.227 \times 10^{-11}$ m$\\$Planck's constant $h = 6.626 \times 10^{-34}$ J s$\\$Mass of electron $m_e = 9.109 \times 10^{-31}$ kg$\\$Charge of electron $e = 1.602 \times 10^{-19}$ C$\\$Substitute these values into the equation for $V: \\ V = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times (1.227 \times 10^{-11})^2} \\$Calculate the numerator:$\\ (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \\$Calculate the denominator:$\\ 2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times (1.227 \times 10^{-11})^2 \\ = 2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times 1.506 \times 10^{-22} \\ = 4.38 \times 10^{-31} \times 1.506 \times 10^{-22} \\ = 6.60 \times 10^{-52} \\$Now, calculate $V: \\ V = \frac{4.39 \times 10^{-67}}{6.60 \times 10^{-52}} \\ = 6.65 \times 10^{4}$ V$\\$Therefore, the potential difference is approximately $10^4$ V, which corresponds to Option 3.