What is the de-Broglie wavelength associated with an electron, accelerated through a potential difference of 10 kV

0.0123 nm

The correct answer is Option (2) → 0.0123 nm

Given:

Potential difference, $V = 10\ \text{kV} = 10^4\ \text{V}$

Charge of electron, $e = 1.6 \times 10^{-19}\ \text{C}$

Mass of electron, $m = 9.11 \times 10^{-31}\ \text{kg}$

Planck's constant, $h = 6.626 \times 10^{-34}\ \text{Js}$

Electron is accelerated through potential difference $V$, its kinetic energy:

$\frac{1}{2} m v^2 = e V$

Velocity: $v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \cdot 1.6 \times 10^{-19} \cdot 10^4}{9.11 \times 10^{-31}}}$

$v = \sqrt{\frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}}} = \sqrt{3.513 \times 10^{15}} \approx 5.93 \times 10^7\ \text{m/s}$

de-Broglie wavelength: $\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 5.93 \times 10^7}$

$\lambda = \frac{6.626 \times 10^{-34}}{5.395 \times 10^{-23}} \approx 1.23 \times 10^{-11}\ \text{m}$

∴ De-Broglie wavelength of the electron = $1.23 \times 10^{-11}\ \text{m}$