Answer
a. The drift speeds are the same.
b. The electrons have less potential energy after leaving the resistor compared to when they entered.
Work Step by Step
a. From equation 25.4, we see that $v_d=\frac{J}{nq}$. The current density J is the same before and after, and the electron concentration and charge stay the same. The drift speed does not change.
b. The electrons lose potential energy, and this energy is transformed into thermal energy within the resistor.
Another way to look at it is that the potential energy is given by qv. Electrons enter the resistor on the side with lower electric potential, which corresponds to higher potential energy (because their charge q is negative).
