What is the theoretical formula for the de-Broglie wavelength of a particle?

The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

$A$ free particle with initial kinetic energy $E$ and de-Broglie wavelength $\lambda$ enters a region in which it has potential energy $V$. What is the particle's new de-Broglie wavelength?

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?