- #1

Metalsonic75

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The nuclei of both 3H and 3He have radii of 1.5 *10^-15 With what velocity must the electron be ejected if it is to escape from the nucleus and not fall back?

I don't quite understand what this question is asking, but given the information (and given that neutrons have no charge) I treated this problem as if there were only two particles: the proton (nucleus) and the electron being ejected. I am unsure if this is the right setup, because of the '3' in front of the 3H and 3He.

I am also unsure of how to calculate this, because the equations we have learned imply that the force acting on a charge is only zero at a "very large distance", ie, the distance is approaching infinity. In theory this means that there is no point where the electron would not be influenced by the proton's force. So how does it escape?

I would really appreciate it if someone could give me a nudge in the right direction.