Скачать или смотреть In an NPN transistor 〖10〗^10 electrons enter the emitter in 〖10〗^(-6) | PGMN Solutions

In an NPN transistor 〖10〗^10 electrons enter the emitter in 〖10〗^(-6) " " s and 2% electrons recombine with holes in base. The current ratios ' α ' and ' β ' of a transistor are respectively (nearly)
(A) 0.98,49
(B) 49.0 .98
(C) 0.49,98
(D) 98,0.49






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🚀 Step-by-Step Solution to Solve This Question 🚀
📌 Chapter: Semiconductor Device
📌 Topic: Current Gain – Alpha (α) and Beta (β) in NPN Transistor

🔹 Step 1: Understand what the question tells us
In this transistor, electrons are flowing from the emitter into the base region. Out of all the electrons entering the base, a small percentage recombines with holes in the base.

🔹 Step 2: Focus on electron movement
If 2% of the electrons recombine in the base, it means 98% of the electrons successfully reach the collector region. This indicates that the collector current is nearly equal to the emitter current.

🔹 Step 3: Analyze alpha (α)
Since alpha is the measure of how efficiently electrons from the emitter reach the collector, and 98% do so, alpha is approximately 0.98.

🔹 Step 4: Analyze beta (β)
Beta shows how much larger the collector current is compared to the base current. Since only 2% of electrons are lost in the base, the base current is very small. This means the collector current is much higher compared to the base current, indicating that beta is large—close to 49 in this case.

✅ Final Answer: (A) 0.98, 49 🚀

🔥 Pro Tip:
When the emitter passes most electrons to the collector and very few to the base, alpha is close to 1 and beta becomes a large number—this is typical in efficient transistor designs!


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