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The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015 cm-3and 5 x 1012 cm-3 respectively. The mobilities of electrons and holes are respectively 23 x 103 cm3 / Vs and 102 cm3 / Vs. Which type of semiconductor is it?
- a)p-type only
- b)n-type only
- c)either n or p type
- d)both n and p type
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The densities of electrons and holes in an extrinsic semiconductor are...
Given data:
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
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Community Answer
The densities of electrons and holes in an extrinsic semiconductor are...
Solution:
Given data:
Density of electrons (n) = 8 x 10^15 cm^-3
Density of holes (p) = 5 x 10^12 cm^-3
Mobility of electrons (μn) = 23 x 10^3 cm^2/Vs
Mobility of holes (μp) = 102 cm^2/Vs
To determine the type of semiconductor, we need to calculate the conductivity type of the extrinsic semiconductor based on the given data.
Conductivity type of the semiconductor can be determined by calculating the majority carrier concentration, which is the carrier type with higher concentration.
Majority carrier concentration (n0) = p for p-type semiconductor, and n for n-type semiconductor
Calculation:
Given that n = 8 x 10^15 cm^-3 and p = 5 x 10^12 cm^-3
Since n > p, the majority carrier concentration is n, so the extrinsic semiconductor is n-type.
Answer: (b) n-type only
Given data:
Density of electrons (n) = 8 x 10^15 cm^-3
Density of holes (p) = 5 x 10^12 cm^-3
Mobility of electrons (μn) = 23 x 10^3 cm^2/Vs
Mobility of holes (μp) = 102 cm^2/Vs
To determine the type of semiconductor, we need to calculate the conductivity type of the extrinsic semiconductor based on the given data.
Conductivity type of the semiconductor can be determined by calculating the majority carrier concentration, which is the carrier type with higher concentration.
Majority carrier concentration (n0) = p for p-type semiconductor, and n for n-type semiconductor
Calculation:
Given that n = 8 x 10^15 cm^-3 and p = 5 x 10^12 cm^-3
Since n > p, the majority carrier concentration is n, so the extrinsic semiconductor is n-type.
Answer: (b) n-type only
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The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015cm-3and 5 x 1012cm-3respectively. The mobilities of electrons and holes are respectively 23 x 103cm3/ Vs and 102cm3/ Vs. Which type of semiconductor is it?a)p-type onlyb)n-type onlyc)either n or p typed)both n and p typeCorrect answer is option 'B'. Can you explain this answer?
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