1. An a-particle (q=+2e)in a nuclear accelerator moves from one terminal at a po

Question

1. An a-particle (q=+2e)in a nuclear accelerator moves from one terminal at a potential of Va=+6.5x10^6 volt to another at a potential Vb=0
i) what is the corresponding change in potential energy of the system?
ii) assuming the terminals and their charges do not move at that no external forces act on the system what is the change in kinetic energy of the particle (in ev)

2. Given deltaV = Vf - Vi where Vi is the potential of the high voltage terminal of a Van de Graaff accelerator and Vf is the potential at the target location

i) calculate the proton kinetic energy, K at the target location when a proton at rest is placed in the high voltage terminal with potential of 10 MV and the potential at the target location is 0

ii) what is the speed of the proton when it strikes the target

An a-particle (q=+2e)in a nuclear accelerator moves from one terminal at a potential of Va=+6.5x10^6 volt to another at a potential Vb=0 what is the corresponding change in potential energy of the system? assuming the terminals and their charges do not move at that no external forces act on the system what is the change in kinetic energy of the particle (in ev) Given delta V = Vf - Vi where Vi is the potential of the high voltage terminal of a Van de Graff accelerator and Vf is the potential at the target location calculate the proton kinetic energy, K at the target location when a proton at rest is placed in the high voltage terminal with potential of 10 MV and the potential at the target location is 0 what is the speed of the proton when it strikes the target

Explanation / Answer

1.
i)dPE = qdV = 2e*(vb - va) = -13.0E6 eV

ii) dKE = - dPE = 13.0E6 eV

2) K = PE = 10.0E6 eV

ii) 10.0E6*1.6E-19 = 0.5*1.6E-27*v^2
v=4.47E7m/s

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