In an NPN transistor 10¹⁰ electrons enter the emitter in 10⁻⁶ ~s and 2 % electrons recombine with holes in base. The current ratios ' ' and ' ' of a transistor are respectively (nearly)

Step-by-step solution

I _ e = n _ e e t and I _ c = n _ c e t From the data given, aligned I_c & = (98 / 100) n_e e t = 98 100 I_c & = I_c I_e = 98 I_e 100 I_e =0.98 aligned = 1- = 0.98 1-0.98 = 0.98 0.02 =49