Question from Atomic Structure

Question

Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm. 

Answer 1

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

   gives:

     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 m_oc² )  

   where KE will be the particle kinetic energy in eV

   and m_oc² 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.