Resistivity Ratios: Germanium Homework Statement

In summary, the conversation discusses the resistivity ratio of n-type Germanium at room temperature and the relative density of vacancies in a solid. It is estimated that the ratio of resistivity between n-type and intrinsic Germanium is given by PExtrinsic/PIntrinsic, with a higher electron hole product resulting in a higher extrinsic resistivity. As for the relative density of vacancies, it is shown that it will always be less than 10-8 per cent unless the melting point is higher than 1000K. The approach for solving this problem involves manipulating the equation n=Ne-E/kT.
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Homework Statement



(a) A sample of n-type Germanium contains 1023 ionized donors per cubic meter. Estimate the ratio at room temperature (300K) of the resistivity of this material to that of high purity intrinsic Germanium. The Electron hole product for Germanium is 1038m-6 and it can be assumed that the mobility of the electrons and the mobility of the electrons in the hole are similar at 300K.

(b) Suppose it takes an energy of 2eV to create a vacany in a certain solid. Show that the relative density of the vacanies of the atoms will always be less than 10-8 per cent unless the melting point is higher than 1000K.

The Attempt at a Solution



Well, I thought since the electron hole product is 1038 then that gives:

Intrinsic - Ne=Nn = 1019
Extrinsic - Ne=1023 therefor Nn=1015

relitive resistivity can be given as PExtrinsic/PIntrinsic.

I'm not really sure how I continue from there. Do I disregard the mobilities because the are similiar?

(b) I'm really unsure how to complete (b), I think it might be something to do with manipulation of n=Ne-E/kT.

Any help would be much appreciated
 
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What is resistivity ratio?

Resistivity ratio is a measure of the resistance of a material to the flow of electric current. It is the ratio of the resistivity of a given material to the resistivity of a standard material, usually copper or silver.

What is the significance of resistivity ratios?

Resistivity ratios are important because they allow us to compare the resistivity of different materials and determine their conductivity. This information is crucial in designing and selecting materials for various electrical and electronic applications.

How is the resistivity ratio of germanium calculated?

The resistivity ratio of germanium is calculated by dividing its resistivity by the resistivity of the standard material (i.e. copper or silver). The resulting value is then compared to the resistivity ratio of other materials to determine its relative conductivity.

What factors can affect the resistivity ratio of germanium?

The resistivity ratio of germanium can be affected by various factors such as temperature, impurities in the material, and crystal structure. Changes in these factors can alter the conductivity of germanium and therefore, its resistivity ratio.

How is the resistivity ratio of germanium used in practical applications?

The resistivity ratio of germanium is used in practical applications such as semiconductor devices, infrared detectors, and solar cells. By understanding the resistivity ratio of germanium, engineers can design and optimize these devices for better performance and efficiency.

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