A proton moves with a velocity equal to $(\frac{1}{20})^{th}$ of velocity of light. The associated de-Broglie wavelength is:

$2.65 × 10^{-14} m$

The correct answer is Option (3) → $2.65 × 10^{-14} m$

Given:

Velocity of proton, $v = \frac{c}{20}$

Mass of proton, $m = 1.67 \times 10^{-27}\ \text{kg}$

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

Speed of light, $c = 3 \times 10^8\ \text{m/s}$

de-Broglie wavelength,

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times (3 \times 10^8 / 20)}$

$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.5 \times 10^7}$

$\lambda = \frac{6.626 \times 10^{-34}}{2.505 \times 10^{-20}} \approx 2.64 \times 10^{-14}\ \text{m}$

∴ de-Broglie wavelength of proton = $2.64 \times 10^{-14}\ \text{m}$