## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) Electron move from higher potential to lower potential against the electric field. (b) $-0.7$ volt
Given: Electron initial speed =5,00,000 m/s Electron final speed =0 (a) Electron move from higher potential to lower potential against the electric field. (b) By conservation of energy; Loss in kinetic energy = Gain in potential energy $e\Delta V_{1}+KE_{i}=e\Delta V_{2}+KE_{f}$ $e\Delta V_{1}+KE_{i}=e\Delta V_{2}+0$ $KE_{i}=e( V_{2}-V_{1})=-e\Delta V$ $-e\Delta V=\frac{1}{2}m_{e}v_{e}^{2}$ $\Delta V=-\frac{{\frac{1}{2}m_{e}v_{e}^{2}}}{1.6 \times 10^{-19}}$ $=-\frac{1}{2}\times \frac{{ 9.11 \times 10^{-31} \times(5\times 10^{5})^{2}}}{1.6 \times 10^{-19}}$ $=-0.7$ volt