1
$\begingroup$

When firing alpha particles head on towards gold nuclei so that they are deflected 180 degrees, the only thing that should affect their deflection is the nuclear charge, the alpha particle's charge, and the initial kinetic energy of the alpha particle, shown by energy = kQq/r where r is the distance of closest approach between the alpha particle and the nucleus.

However, alpha particles not approaching the nucleus directly are deflected at an angle - is this angle related to the size or mass of the nucleus at all? Would two nuclei with equal charges cause equal deflection if one nucleus had twice as many neutrons and so was much larger and heavier?

My textbook states that Rutherford reasoned "the core of the atom must be massive on an atomic scale to deflect alpha particles through large angles", implying the angle of deflection is affected by the mass of the nucleus, however it later states that increasing the number of neutrons in the nucleus would not affect the deflection as the charge is the same.

$\endgroup$
3
$\begingroup$

The usual derivation of the differential scattering cross section makes the assumption that the mass of the target nucleus is much greater than that of the incoming alpha particle. This is saying that the nucleus does not recoil when it interacts with alpha particle. Better still assume that the $\csc^4$ formula derivation is done in the centre of mass frame of the nucleus and alpha.
Since observations are made in the laboratory frame a correction has to be applied to the $\csc^4$ formula and that correction does depend on the mass of the nucleus in relation to the mass of the alpha and so changing the number of neutrons whilst keeping the number of protons the same does change the angle of deflection.
The smaller the nucleus the greater the correction term when the number of neutrons changes.


There are many Internet sites and textbooks which derive the correction term and here is one with the centre of mass derivation shown on the previous page.

$\endgroup$
  • 2
    $\begingroup$ As a side note the identification of alpha particles with helimu nuclei comes from noticing the distictive right-angle between the tracks of alphas and the recoiling nuclei when letting alpha impinge on helium gas. This is a special result for the case when the incident and scattered particles have the same mass. $\endgroup$ – dmckee Apr 5 '18 at 22:12
  • 1
    $\begingroup$ Just going into a center-of-mass frame makes it clear that the mass of both nuclei have something to do with the problem... $\endgroup$ – Jon Custer Apr 5 '18 at 22:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.