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Helga [31]
3 years ago
12

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

at room temperature: σ (????∙m) –1 n (m–3 ) p (m–3 ) Intrinsic 8.9 × 10^4 8.7 × 10^23 8.7 × 10^23Extrinsic 2.3 × 10^5 7.6 × 10^22 1.0 × 10^25 (p-type)Calculate electron and hole mobilities.

Engineering
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

\sigma = n|e|\mu_e+p|e|\mu_h (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

\mu_e= mobility of electron

\mu_h=mobility of holes

\sigma = electrical conductivity

Making the substitution in (S)

Mobility of electron

8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)

0.639=\mu_e+\mu_h

Mobility of hole in (S)

2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))

0.1436 = 7.6*10^{-3}\mu_e+\mu_h

Then, solving the equation:

0.639=\mu_e+\mu_h (1)

0.1436 = 7.6*10^{-3}\mu_e+\mu_h (2)

We have,

Mobility of electron \mu_e = 0.5m^2/V.s

Mobility of hole is \mu_h = 0.14m^2/V.s

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A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

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Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

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Now,

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8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

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Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

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we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
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Answer:

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R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25

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