A potential barriers of 0.5 eV exists in p-n junction. If an electron approaches the p-n junction from the n – side with speed of 5×105m/s ,then electron will enter the P side with velocity (in m/s)

Let electron enters the n region with speed v1 and enters the P region with speed  v2Applying conservation of energy,                                                         12mv12=eV+12mv22⇒12×9.1×10−31×(5×105)2=1.6×10−19×0.5+12×9.1×10−31v22⇒113.75×10−21=0.8×10−19+12×9.1×10−31v22⇒4.55×10−31v22=113.75×10−21 - 80×10−21⇒v22=33.75×10−214.55×10−31⇒v22 = 7.41×1010⇒v2 = 2.7×105⇒v2=27×104m/s