Two identical pure germanium crystals, A and B , are doped with...

Crystal A is doped with arsenic (a pentavalent impurity), making it an n-type semiconductor where electrons are the majority charge carriers.

Crystal B is doped with indium (a trivalent impurity), making it a p-type semiconductor where holes are the majority charge carriers.

The electrical conductivity of a semiconductor is given by $\sigma = e(n_e \mu_e + n_h \mu_h)$.

For the n-type crystal A, the conductivity is mainly due to electrons: $\sigma_A \approx e n_e \mu_e$.

For the p-type crystal B, the conductivity is mainly due to holes: $\sigma_B \approx e n_h \mu_h$.

It is given that the number density of majority carriers in both crystals is the same, so $n_e = n_h$.

Since the mobility of electrons is greater than the mobility of holes ($\mu_e > \mu_h$), it follows that $\sigma_A > \sigma_B$.

Answer: $\sigma_A > \sigma_B$