A positron is moving at 1.0x10^7 m/s in a positive direction of an x axis when,

Question

A positron is moving at 1.0x10^7 m/s in a positive direction of an x axis when, at x=0 it encounters an electirc field directed along the x axis. The electric potential V associated with the field is given by the graph (which I can't upload) that looks like this... From (x(cm) = 0, V=0) to (x=20, V=500) platues to (x=50, V=500) and drops vertically to (x=50,V=500). (a) does the positron emerge from the field at x=0 (direction reversed) or at x=.50m (direction not reversed)? (b) What is its speed when it emerges?

Explanation / Answer

The easiest way to do this is to do the energy in electron volts. 1 eV = 1.602 * 10-19 J.

This is because we can see the positron is working uphill against a linearly increasing potential whose peak is 500 V; so unless the positron's incoming energy is greater than 500 eV, it will have to reverse.

K = ½mv2 = (0.5)(9.11 * 10-31 kg)(1.0 * 107 m/s)2 = 4.555 * 10-17 J = 284.3 eV.

(a) So we can see it will reverse.

(b) 1.0 * 107 m/s. This is an elastic collision, it rebounds as fast as it goes in, just in the opposite direction.

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