$v = 1500$ m/s

A particle’s de Broglie wavelength can be computed as follows:

$$
begin{align*}
lambda &= dfrac{h}{p} \
&= dfrac{h}{m_{proton} cdot v} \
&= dfrac{6.626 times 10^{-34} text{ Js}}{(1.67 times 10^{-24} text{ kg}) times (1500 text{ m/s})} \
&= boxed{2.64 times 10^{-13} text{ m}}
end{align*}
$$

We know that the relationship between the momentum of a particle $p$ and the wavelength associated with it $lambda$ is given by

$$
begin{align}
lambda=frac{h}{p}
end{align}
$$

where $h=6.62times 10^{-34}text{Js}$ is Planck’s constant.

The kinetic energy $K$ of the electron is:

$$
begin{align*}
K&=5.5text{eV}\
&=5.5times 1.6 times 10^{-19} text{J}\
&=8.8times 10^{-19}text{J}
end{align*}
$$

Nonrelativistic expression for the kinetic energy of an electron $K$ in terms of its momentum $p$ is given by

$$
begin{align*}
K=frac{p^2}{2m}
end{align*}
$$

where $m=9.11times 10^{-31}text{kg}$ is the mass of the electron. From this equation, we can express the momentum $p$

$$
begin{align*}
p=sqrt{2mK}
end{align*}
$$

By replacing $p$ into Eq. (1), we get

$$
begin{align*}
boxed{lambda}&=frac{h}{p}\
&=frac{h}{sqrt{2mK}}\
&=frac{6.62times 10^{-34}text{Js}}{sqrt{2times 9.11 times 10^{-31}text{kg} times 8.8times 10^{-19}text{J}}}\
&=5.5times 10^{-10}text{m}\
&=boxed{0.55text{nm}}
end{align*}
$$