Forces at Play in Geiger-Marsden Alpha Particle Scattering

AI Thread Summary
The Geiger-Marsden experiment demonstrates that the scattering of an alpha particle by a gold atom is primarily due to the electrostatic force between the alpha particle and the nucleus of the gold atom, confirming option B as correct. The discussion clarifies that while the nuclear force, mentioned in option D, refers to the strong interaction between nucleons, it is not the primary force responsible for the scattering in this context. The strong nuclear force is attractive at larger distances but becomes repulsive at very short ranges. Understanding the nuclear force requires knowledge of quantum chromodynamics (QCD), which may be beyond the scope of basic physics studies. Overall, the key takeaway is that the electrostatic force plays a crucial role in alpha particle scattering.
lingling
Messages
22
Reaction score
0

Homework Statement



Which of the following forces is responsible for the scattering of an alpha particle by a gold atom in the Geiger-Marsden experiment?

A. electrostatic force between the alpha particle and the electrons of the gold atom

B. Electrostatic force between the alpha particle and the nucleus of the gold atom.

C. Gravitational force between the alpha particle and the nucleus of the gold atom

D. Nuclear force between the alpha particle and the nucleus of the gold atom.

Homework Equations



nil

The Attempt at a Solution



I 've chosen B and this is correct.
But I don't know what 's the meaning of 'nuclear force' in option D.
 
Physics news on Phys.org
The nuclear force was considered to be one of the fundamental forces and can be thought of to act between nucleons. However, the 'nuclear force' is now understood in terms of the strong interaction (in QCD) which acts between gluons, quarks and anti-quarks, however I feel this is a little deeper than you need to go. Unless you are studying Physics at Degree level, then all you need to know is that the strong nuclear force acts between nucleons and is strongly attractive, except at small distances where it becomes strongly repulsive.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top