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On applying an electric field of intensity 10 V/cm across a semiconductor at a certain temperature the average drift velocity of free electrons is measured to be 70 m/s. Then the electron mobility is
- a)7 × 104 cm2/Vs
- b)700 cm2/Vs
- c)7 cm2/Vs
- d)700 cm/Vs
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
On applying an electric field of intensity 10 V/cm across a semiconduc...
Concept:
Mobility: It denotes how fast is the charge carrier is moving from one place to another.
It is demoted by μ.
Mobility is also defined as:
Where,
Vd = Drift velocity
E = field intensity
Calculation:
E = 10 V/cm
Vd = 70 m/s = 7000 cm/s
From equation (1):
μ = 7000/10
μ = 700 cm2/Vs
Note:
Electron mobility is always greater than hole mobility,
Hence electron can travel faster and contributes more current than a hole.
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Community Answer
On applying an electric field of intensity 10 V/cm across a semiconduc...
The electron mobility (μ) can be calculated using the equation:
μ = v_d / E
where μ is the electron mobility, v_d is the average drift velocity, and E is the electric field intensity.
Given:
v_d = 70 m/s
E = 10 V/cm = 100 V/m (since 1 V/cm = 100 V/m)
Substituting the given values into the equation:
μ = 70 m/s / 100 V/m = 0.7 m^2/Vs
Therefore, the electron mobility is 0.7 m^2/Vs.
μ = v_d / E
where μ is the electron mobility, v_d is the average drift velocity, and E is the electric field intensity.
Given:
v_d = 70 m/s
E = 10 V/cm = 100 V/m (since 1 V/cm = 100 V/m)
Substituting the given values into the equation:
μ = 70 m/s / 100 V/m = 0.7 m^2/Vs
Therefore, the electron mobility is 0.7 m^2/Vs.
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