Just so that all your questions are answered I looked into this...
I'm not sure what basic equations (or their form or units) you are using...
Most basically and in mks units we have
wavelength = h [plancks const] / p [particle momentum]
putting in desired wavelength = 5 x10-12 meters (0.005 nm)
and h = 6.626 x10-34 m2kg/s
p = 1.33 x10-22 kgm/s [mass times velocity which is a momentum]
Substituting v = p/m into E = (1/2)mv2 gives
E = p2/(2m)
We have p , and m for proton is 1.673 x10-27 kg, so we get
E = 5.288 x10-18 kg m2/s2 [aka Joules]
Convert this energy into "electron volts" [energy a 1-electron charge
gains going through 1 V potential] by dividing:
E[eV] = 5.288 x10-18 Joules / 1.602 x10-19 Joules [ = 1 eV]
= 33 eV
Since the proton has a charge of 1,
the accelerating potential needed is 33 V.
- - - - - - -
particle wavelength = (hc)/(pc)
where (hc) is Planck's constant times the speed of light;
these together have units of energy times length,
a useful combination/units gives 1239.84 eV[energy] nm[length]
We want wavelength = 0.005 nm so combined with the above constant
we get that we want:
(pc) = 2.48 x105 eV
The pc of the particle is related to its kinetic energy and rest mass through:
pc = sqrt( 2 KE moc2 )
where KE will be the particle kinetic energy in eV
and moc2 is the particle rest mass in eV.
and pc will then also be in eV.
For a proton the rest mass is 938 Mev = 9.38 x108 eV
So putting in the desired pc value from above and squaring and solving for KE we get:
KE = 6.15 x1010 / (2 * 9.38 x108)
= 32.8 eV
The eV unit of energy is the amount of energy a 1-electron charge acquires when
it passes through a 1 V potential. Since the proton has the same charge as an
electron, the proton will acquire 32.8 eV of energy by an accelerating potential
of 32.8 V.