Solving Problems Involving Ge, Well Lagged Wires, and Heat Flow

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SUMMARY

This discussion addresses two specific problems involving Germanium (Ge) and heat flow in well-lagged wires. In the first problem, it is established that 30% of the current in a Ge sample is carried by electrons, given their mobility of 3500 cm2/V.s compared to holes at 1500 cm2/V.s. The second problem involves calculating the rate of heat flow along a wire with thermal conductivity defined as K = B + CT, leading to the formula dQ/dt = A/L(T1-T2) {B + C/2 (T1 + T2)}. Key principles include Hall effect measurements and Fourier's law of heat conduction.

PREREQUISITES
  • Understanding of Hall effect measurements in semiconductors
  • Knowledge of thermal conductivity and Fourier's law
  • Familiarity with electrical mobility concepts in materials
  • Basic principles of heat transfer in one-dimensional systems
NEXT STEPS
  • Study Hall effect calculations in semiconductor physics
  • Explore Fourier's law of heat conduction in detail
  • Investigate the relationship between mobility and current in semiconductors
  • Learn about thermal conductivity variations with temperature in materials
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Students and professionals in physics, electrical engineering, and materials science who are dealing with semiconductor behavior and thermal analysis in conductive materials.

tigigi
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Could anyone gives me some hints of how to solve these problems ?

1. a sample of Ge has both electrons and holes carrying the current. When a hall measurement is done, there's no hall effect. If the mobility of electrons in Germanium is 3500 cm^2/V.s and that of holes is 1500 cm^2/V.s, show that 30% of the current is carried the electrons.

2. a well lagged wire (no heat enters or leaves except at the ends) length L and cross sectional area A has its ends maintained at T1 and T2. The thermal conductivity of the wire K=B+CT where B and C are constants. Show that the rate of flow of heat along the wire is given by dQ/dt = A/L(T1-T2) {B+C/2 (T1+T2)}

Thank you. I appreciate it.
 
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For 1, can one convert mobility into current.

Incidentally 1500/(1500 + 3500) = 0.3 or 30%, but 3500 cm^2/V.s is the mobility of electrons and 1500 cm^2/V.s the mobility of holes, so the electrons move much more readily.

for 2, think about Fourier's law in one dimension.

dQ/dt = - k A dT/dx

and remember k = k(T)
 

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