EXAMPLE DATA: Beam: K - ; E = 4,2 GeV ; m K =m p /2 Target: Hydrogen atoms; R nucleus =10 -5 R atom me = mp/2000.

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Presentation transcript:

EXAMPLE DATA: Beam: K - ; E = 4,2 GeV ; m K =m p /2 Target: Hydrogen atoms; R nucleus =10 -5 R atom me = mp/2000

QUESTIONS 1.Draw a diagram showing the tracks formed by different charged particles. 2.What is the direction of the magnetic field in this picture?. Notice: negative particles such as electrons turn clockwise and positive particles such as protons turn anticlockwise in this magnetic field. 3.Explain the interactions observed in the picture. 4.What can we tell about the energy transferred in these interactions?. 5.Consider the collision particle K - and proton. Compare the radius of curvature of these particles’ trajectories.

ANSWERS 1.This coloured diagram shows the paths followed by these charged particles. Code: green K - ; red electron ; blue proton.

2.The direction of the magnetic field is towards axe X negative. We use the formula of Lorentz force to calculate the value and direction of the force exert on charged particles by a magnetic field. The orientation of, and can be also found using RHR right hand rule or corkscrew rule. X Y Z

3.In point A, the K - particle hits an electron and this electron makes its own characteristic curly track.  In point B, the K - strikes a proton. After the collision the proton moves to the right causing the dark path shown on the picture. 4.  The charged particles passing along the Buble Chamber deposit energy for ionization of Hydrogen atoms. This amount of energy (13,6 eV) is negligible compared with the energy of the incoming particles K - (4,2 GeV).  On the interaction referred in point A, there are more energy transferred to the electron. This curly track requires an energy about MeV’s.  On the collision between K - and proton, the proton takes some energy from the K -. This proton loses its energy as it moves slowly creating a short dark path.

We consider that the forces exert on charged particles by the magnetic field inside a BC are centripetal forces. We use it to know the relation between the momentum of the particle and the radius of curvature of its trajectory. m v 2 / R = q v B p = R q B We can conclude that shorter radius (or more curvature ) indicates lower momentum. conclusion: R p < R K- and p p < p K-