The wavelength of the de-Broglie wave associated with an electron accelerated through 150 V is:

1 Å

The correct answer is Option (1) → 1 Å

de-Broglie wavelength: $\lambda = \frac{h}{p}$

Electron accelerated through potential $V$: kinetic energy $K.E. = eV = \frac{1}{2} m v^2$

Momentum: $p = \sqrt{2 m e V}$

Given: $V = 150 \text{ V}$, $m = 9.11 \times 10^{-31} \text{ kg}$, $e = 1.6 \times 10^{-19} \text{ C}$, $h = 6.63 \times 10^{-34} \text{ Js}$

$p = \sqrt{2 \cdot 9.11 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 150} \approx \sqrt{4.37 \times 10^{-47}} \approx 6.61 \times 10^{-24} \text{ kg·m/s}$

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} \approx 1.0 \times 10^{-10} \text{ m} = 0.1 \text{ nm}$

Answer: $\lambda \approx 0.1$ nm