Question

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​

Answer 1

sorry I don't know it's answer

Answer 2

Answer:

The mobility of the charge carriers, $\mu_{e}$ measured is $14.7\times 10^{-3}m^{2}V^{-1}s^{-1}$.

Therefore, option a) $14.7\times 10^{-3}m^{2}V^{-1}s^{-1}$ is correct.

Explanation:

Given,

The hall coefficient of the given specimen, $R_{h}$ = $-7.35\times 10^{-5} m^{3}C^{-1}$

The conductivity of the specimen, $\sigma_{e}$ = $200\Omega^{-1}m^{-1}$

The nature of the semiconductor =?

The mobility of the charge carriers, $\mu_{e}$ =?

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 = $\sigma_{e}\times R_{h}$
• Mobility = $200\times 7.35\times 10^{-5}$ = $14.7\times 10^{-3}m^{2}C^{-1}\Omega^{-1}$

Also,

• $C^{-1}\Omega^{-1} = V^{-1} s^{-1}$

Therefore,

• Mobility = $14.7\times 10^{-3}m^{2}V^{-1}s^{-1}$.