Answer
(a) Electron move from higher potential to lower potential against the electric field.
(b) $-0.7$ volt
Work Step by Step
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
