4.53

(a) An electron (charge -e and mass m) is in a circular orbit of radius r around a fixed proton (charge +e). Prove that the electron's KE is equal to -1/2 times its PE; that is, T= -U/2 and hence E= U/2.

$\text{From }f_{c}=\frac {ke^{2}}{r^{2}} \text{ and } F_{inward} = F_{outward}\\\\ \frac {ke^{2}}{r^{2}}=\frac {mv^{2}}{r}\\\\ mv^{2}=\frac{ke^{2}}{r}\\\\ Total\;energy : E= T + U\\\\ E = \frac{1}{2}mv^{2} + (\frac{-ke^{2}}{r})\\\\\therefore E = \frac{1}{2}\frac{-ke^{2}}{r} + \frac{-ke^{2}}{r}\\\\ U = \frac{-ke^{2}}{r} \;\;and\;\; T=-\frac{1}{2}U$

(b) Inelastic collision : Electron number 1 is in a circular orbit of radius r around a fixed proton.Electron 2 approaches from afar with kinetic energy $T_{2}.$ When the second atom hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius $r^\prime$ . Write down the expression for the total energy of the three-particle system in general.

$Total\;energy : E= T_{1} + T_{2} + U_{1} + U_{2} + U_{1by2}\\\\ \therefore E = \frac{1}{2}mv^{2}_{1} + \frac{1}{2}mv^{2}_{2} + \frac{-ke^{2}}{r_{1}} + \frac{-ke^{2}}{r_{2}} + \frac{-ke^{2}}{r_{1by2}}$

(c) Identify the value of all five terms and the total energy E long before the collision occurs and long after it is over. What is the KE of the outgoing electron 1 once it is far away?

$Long\;before\; the\; collision\\\\ E= T_{1} + T_{2} + U_{1}\\\\ E = -\frac{1}{2}U_{1} + T_{2} + U_{1} \;\; = T_{2}+ \frac{1}{2}U_{1}\\\\ \therefore E = T_{2} - \frac{ke^{2}}{2r}\\\\ Long\;after\;the\collision\\\\ E^\prime = T^\prime_{1} + T_{2}^\prime + U^\prime_{2}\\\\ \therefore E^\prime = T^\prime_{1}-\frac{ke^{2}}{2r^\prime}\\\\ Once\;the\;electron1\;is\;far\;away\\\\ \text{Conservation of energy: } E = E^\prime\\\\ T_{2} - \frac{ke^{2}}{2r} = T^\prime_{1}-\frac{ke^{2}}{2r^\prime}\\\\ \therefore T^\prime_{1}= T_{2} +\frac{ke^{2}}{2r^\prime} - \frac{ke^{2}}{2r}$

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