Dan D. answered • 05/23/16

Patient Tutor Focused on Your Understanding of Math

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}m^{2}kg/s gives:

p = 1.33 x10

^{-22}kgm/s [mass times velocity which is a momentum] Substituting v = p/m into E = (1/2)mv

^{2}gives E = p

^{2}/(2m) We have p , and m for proton is 1.673 x10

^{-27}kg, so we get E = 5.288 x10

^{-18}kg m^{2}/s^{2 }[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 x10

^{5}eVThe pc of the particle is related to its kinetic energy and rest mass through:

pc = sqrt( 2 KE m

_{o}c^{2}) where KE will be the particle kinetic energy in eV

and m

_{o}c^{2}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 x10

^{8}eVSo putting in the desired pc value from above and squaring and solving for KE we get:

KE = 6.15 x10

^{10}/ (2 * 9.38 x10^{8}) = 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.