Question

In a sample of Ge at room temperature (293 K), what fraction of the Ge atoms must be replaced with donor atoms in order to increase the electron population in the CB by 3X? Assume all donor atoms are ionized, and take the energy gap in Ge to be 0.66 eV.

Answer 1

EXPLAINATION :

Ouven Temperature = 293k Energy gap (Eg) = 0.66er Now we know that law of mass action: mp (RT) (met mi) e . I? Tha Now, in the case of * without doping n=p n = P = √12-11 with doping N-P=D ( is doping? white law of mass action still holds sp= I2 this gields Y nap + (np) ²+ 40P - Dt 0² +4 12 if after doping n = 3I - D+ √² +972 6 I-D - Dr. 4 I 2 in solning I2 - 12 ID [D-diſ

mee = 0.56 me and mh - 0.29 me along Eg = obser, we get 2 -3 I- 1.31x 10 cm Heuce D-8I 35x10 cons Now we know that ere has & atoms per unit cell its lattice constant -5.654. Number density of Gre atom N = (5.45 x 35 cm) 5.65 In N= 4.44 x 10cm3 22 e. So the required fraction is = 7.9 x 10-10 Aus