I will set the wavelength of the electron to be equal to the diameter of the protein molecule:
$sf(lambda=2.4xx10^(-9)color(white)(x)m)$
The de Broglie expression gives us:
$sf(lambda=h/(mv))$
$:.$$sf(v=h/(mlambda))$
$sf(v=(6.63xx10^(-34))/(9.11xx10^(-31)xx2.4xx10^(-9))color(white)(x)"m/s")$
$sf(v=3.03xx10^(5)color(white)(x)"m/s")$
Kinetic energy $sf(=1/2mv^2)$
$:.$$sf(KE=1/2xx9.11xx10^(-31)xx(3.03xx10^5)^2color(white)(x)J)$
$sf(KE=4.18xx10^(-20)color(white)(x)J)$