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san4es73 [151]
3 years ago
5

A cube that is 20 nanometer on an edge contains 399,500 silicon atoms, and each silicon atom has 14 electrons and 14 protons. In

the silicon we replace 4 silicon atoms with phosphorus atoms (15 electrons and 15 protons/atom), and we replace 7 silicon atoms with boron atoms (5 electrons and 5 protons/atom). How many "holes" are available to carry current at 300K? Holes look like positive mobile carriers. Three significant digits and fixed point notation.
Physics
1 answer:
Sergeu [11.5K]3 years ago
6 0

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

  1. N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.
  2. P-type Impurities: Trivalent impurities e.g. Boron, Aluminium are added to have a positive "hole" in the structure. Thus a trivalent impurity creates 1 hole.

Now for estimation of extra electrons in the impured structure is as

N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\

Now for estimation of "holes"  in the impured structure is as

N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3

So total 3 "holes" are available for conduction of current at 300K.

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Answer:

7.48 x 10⁵ m

Explanation:

g = 7.86 N/kg

M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m.

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g = GM/(R + h)²

(R + h)² = GM/g = 6.67 x 10⁺¹¹ x 5.97 x 10²⁴ /7.86 = 5.066 x 10¹³

R + h = 7.12 x 10⁶ m

so

h =  7.12 x 10⁶ - 6.37 x 10⁶ = 7.48 x 10⁵ m

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A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.

Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.

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0 = 7 − 4.9 t²

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6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²

v₀ ≈ 5.27 m/s

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