The Hall coefficient of certain specimen was found to be -7.35 x 10-5m2C-1 from 100K to
400K. Determine the nature of semiconductor if the conductivity was found to be 200 Ω
-1m-1
.
Calculate the mobility of the charge carriers.
a.14.7 x 10-3m2V
-1
s
-1
b. 1.47 x 10-3m2V
-1
s
-1
c. 34 x 10-3m2V
-1
s
-1
d. 3.4 x 10-3m2V
-1
s
-1
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Answer:
The mobility of the charge carriers, measured is .
Therefore, option a) is correct.
Explanation:
Given,
The hall coefficient of the given specimen, =
The conductivity of the specimen, =
The nature of the semiconductor =?
The mobility of the charge carriers, =?
Now, as given, the hall coefficient value is negative.
Thus, the negative sign indicates the type of semiconductor.
- In an n-type semiconductor, the majority of carriers are free electrons.
- They accumulate at the bottom surface of the semiconductor and produce a negative charge on the bottom surface.
- Thus, the hall voltage produced in the n-type semiconductor is negative.
Therefore, the negative sign indicates that the nature of the semiconductor is n-type.
Now, we have to find out the mobility of charge carriers:
- Mobility =
- Mobility = =
Also,
Therefore,
- Mobility = .
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