An alpha particle (the nucleus of a helium atom, with charge +2e and a mass four

Question

An alpha particle (the nucleus of a helium atom, with charge +2e and a mass four times that of a proton) and an antiproton (which has the same mass as a proton but charge -e) are released from rest a great distance apart. They are oppositely charged , so each accelerates toward the other. What are the speeds of the two particles when they are 2.5 nm apart?

HINT: You'll need to use two conservation laws. And what does "a great distance" suggest about the initial value of r?

Explanation / Answer

we will use conservation of energy and conservation of momentum laws. At the beginning, the system has nearly 0 potential electrical energy, because of great distance. As well, kinetic energy of the system also 0, because both particles are in rest. The Elect. potential energy of the He at the point requires, is EP=q*V, when V is the potential due to the proton field at distance 2.5nm form the proton Potential due to proton, V=k*q'/R, q'- antiproton charge. V=-0.576 V then E=q*V=2*1.6e-19*0.576=-1.8432e-19 J, so He have lost potential energy, this loss is transferred to the kinetic energy: m*v^2/2=E --> v^2=2*E/m, v^2=2*1.8432e-19/ 6.68e-27 --> v=7.45e3 m/sec the speed of antiprotom we will calculate from the moment conservation m1v1=m2v2 --> v2=m1*v1/m2=v1*4=2.98e4 m/s (m1/m2=4)

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