An electron is accelerated through a potential difference of 36 V. The de-Broglie wavelength associated with it will be

$2 × 10^{-10}m$

The correct answer is Option (1) → $2 × 10^{-10}m$

Energy gained by the electron: $eV = \frac{1}{2} m v^2$

Given: $V = 36 \ \text{V}$, $e = 1.6 \times 10^{-19} \ \text{C}$, $m = 9.11 \times 10^{-31} \ \text{kg}$

Velocity of electron:

$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \cdot 1.6 \times 10^{-19} \cdot 36}{9.11 \times 10^{-31}}}$

$v = \sqrt{\frac{1.152 \times 10^{-17}}{9.11 \times 10^{-31}}} = \sqrt{1.264 \times 10^{13}} \approx 3.556 \times 10^6 \ \text{m/s}$

de-Broglie wavelength:

$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 3.556 \times 10^6}$

$\lambda \approx 2.05 \times 10^{-10} \ \text{m} = 0.205 \ \text{nm}$

de-Broglie wavelength ≈ 0.205 nm