Pure silicon at 300 K has equal electron and equal hole concentration of 1.5 x \( {10 }^{16 } \) \( { m}^{ -3} \) doping by indium increases the hole concentration to 4.5 x \( { 10}^{22 } \) \( { m}^{ -3} \). Then the electron concentration in doped silicon is (in \( { 10}^{ -3} \))
Answer & explanation
Correct answer: option 2
ne x nh = ni2
ni = 1.5 x \( {10 }^{16 } \) \( { m}^{ -3} \)
nh = 4.5 x \( { 10}^{22 } \) \( { m}^{ -3} \)