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The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
A
current in p-type > current in n-type.
B
current in n-type > current in p-type.
C
No current will flow in p-type, current will only flow in n-type.
D
current in n-type = current in p-type.
Detailed Solution
In n-type the majority carriers are electrons; in p-type they are holes.
$I = neAv_d = neA(\mu E)$
Since electron mobility is greater than hole mobility ($\mu_e > \mu_h$), $I_e > I_h$.
So current in n-type > current in p-type.