A neutron initially at rest, decays into a proton, an electron, and...

By the principle of conservation of linear momentum, the total initial momentum of the neutron is zero. Thus, the vector sum of the momenta of the decay products must be zero.

$\vec{p}_p + \vec{p}_e + \vec{p}_\nu = 0$

$\vec{p}_p = -(\vec{p}_e + \vec{p}_\nu)$

Since the electron and antineutrino are ejected along perpendicular directions, the magnitude of the proton's momentum is given by:

$p_p = \sqrt{p_e^2 + p_\nu^2}$

Given:
$p_e = 1.4 \times 10^{-26} \text{ kg-m/s} = 14 \times 10^{-27} \text{ kg-m/s}$
$p_\nu = 6.4 \times 10^{-27} \text{ kg-m/s}$

Substituting the values:

$p_p = \sqrt{(14 \times 10^{-27})^2 + (6.4 \times 10^{-27})^2}$

$p_p = \sqrt{196 + 40.96} \times 10^{-27}$

$p_p = \sqrt{236.96} \times 10^{-27} \approx 15.39 \times 10^{-27} \text{ kg-m/s}$

The recoil speed of the proton is:

$v_p = \dfrac{p_p}{m_p}$

$v_p = \dfrac{15.39 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 9.2 \text{ m/s}$