MHT CET202122 Sep 2021Morning ShiftPhysicsSemiconductorsActual
In an n-p-n transistor 200 electrons enter the emitter in 10⁻⁸ second. If 1 % electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [ e =1.6 10⁻¹⁹ C ]
Options
- A2 10⁻¹⁰ ~A and 49
- B3.2 10⁻⁹ ~A and 99
- C1.6 10⁻¹⁹ ~A and 90
- D1.7 10⁻¹¹ ~A and 70
Correct answer
B. 3.2 10⁻⁹ ~A and 99
Step-by-step solution
aligned & q =200 1.6 10⁻¹⁹ C , t =10⁻⁸ ~s & Emitter current I _ e = q t = 3.2 10⁻¹⁷ 10⁻⁸ =3.2 10⁻⁹ ~A & I _ b = 1 100 I _ e & I _ c = 99 100 I _ e aligned Current amplification factor = I_c I_b =99