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categoryفيزياء
schoolبكالوريوس
event_available2026-07-13
السؤال
Transcribed Image Text:
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Question 2 (35 Points)
One can conceive an atomic nucleus, as a sphère charged uniformly, and
positively. Let the proton number be Z; the charge intensity of either electron
or proton is e -1.6x10-19 Coulomb. Thereby the charge intensity of the nucleus
at hand, becomes Ze. Let its radius be R.
a) (5 p) Consider a location denoted by the distance r to the center of the
nucleus, such that r>R, and write at r (in MKS unit system), the electric field
intensity E(r) and the electric potential V(r) created by the nucleus.
b) (5 p) Write the relationship between these two quantitites, and the reason
for it.
c) (10 p) In Rutherford experiment, helium nuclei also known as alpha particles,
which consist of two protons and two neutrons were accelerated through an
electric potential difference, and then directed towards a very thin Gold foil.
Although most of these charged particles passed through the foil, almost
unaffectedly, some were observed to be deflected sharply, almost backwards.
Rutherford (1909 -1911) was thereby led to the conclusion that Gold atoms
consist mostly of void, but have strongly and positively charged central
regions, namely nuclei.
In such an experiment assume that an alpha particle with a positive charge 2e
is accelerated through a potential difference of AV and then happens to strike,
head on, a Gold nucleus of positive charge Ze. The energy W it would then
acquire would be (by definition of electric potential) W-2еAV. The distance of
closest approach d, in such a special collision is expected to give an
approximate value for the radius of gold nucleus. First, show that the alpha
electric potential energy W, at d (in MKS unit system), becomes
W =
2Ze
4x.d.
and thereby, one can write for the closest distance di
Ze
d=
4 AV
d) (10 p) Calculate d, in the light of the following data, in terms of the radius
R of the Gold nucleus, which had been afterwards measured.
1
4π
-=9x10'N m²/C2 (MKS), AV-6 x 106 Volt, Z-79, R-8x10-15 m.
e) (5 p) Given that alpha's mass is m = 6.6 x 10-27 kg, find its velocity in
meters/ second.
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