Answer
$42.82$
Work Step by Step
Let's assume that the voltage of acceleration is $V$.
We know that two particles are accelerated from rest, so
their kinetic energy is given by
$$KE=\frac{1}{2}mv^2=eV$$
Thus, the speed is given by
$$v=\sqrt{\dfrac{2eV}{m}}\tag 1$$
According to de Broglie, the wavelength of the particle is given by
$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$
Plugging from (1)
$$\lambda=\dfrac{h}{m\sqrt{\dfrac{2eV}{m}} }$$
$$\lambda=\Delta x=\dfrac{h}{\sqrt{ 2meV } }\tag 2$$
We know that
$$\Delta p_x\Delta x\geq \hslash$$
Thus,
$$\Delta p_x\Delta x= \hslash=\dfrac{h}{2\pi }$$
Therefore,
$$ \Delta p_x =\dfrac{h}{2\pi \Delta x } $$
Plugging from (2);
$$ \Delta p_x =\dfrac{h}{2\pi \dfrac{h}{\sqrt{ 2meV } } } $$
$$\boxed{ \Delta p_x =\dfrac{\sqrt{ 2meV} }{2\pi } }$$
Now we need to find the ratio of the uncertainty of the proton's momentum and the electron's momentum.
$$\dfrac{\Delta p_p}{\Delta p_e}=\dfrac{\dfrac{\sqrt{ \color{red}{\bf\not}2m_p\color{red}{\bf\not}e\color{red}{\bf\not}V} }{\color{red}{\bf\not}2\color{red}{\bf\not}\pi }}{\dfrac{\sqrt{ \color{red}{\bf\not}2m_e\color{red}{\bf\not}e\color{red}{\bf\not}V} }{\color{red}{\bf\not}2\color{red}{\bf\not}\pi }}$$
$$\dfrac{\Delta p_p}{\Delta p_e}=\sqrt{\dfrac{m_p}{m_e}}$$
Plugging the known;
$$\dfrac{\Delta p_p}{\Delta p_e}=\sqrt{\dfrac{1.67\times10^{-27}}{9.11\times10^{-31}}}=\color{red}{\bf 42.82}$$
