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3 years ago

how does the location of semiconductors on the periodic table differ from the locations of other element families

Physics

1 answer:

Mekhanik [1.2K]3 years ago

3 0

Answer:

Semi conductors are part of the fourth group in periodic table.

Explanation:

Periodic table is the arrangement of chemical elements on the basis of their atomic number. So all the elements with respective valence electrons are placed in thir respective groups. . All the elements with valence electrons four are placed in the fourth group .Semi conductors is the common name given to the elements of 4th group. These elements dont have the tendency to either donate or accept.

The group 3 elements have three electrons in their valence shell and hence have a tendency to donate. Whereas the group 5 elements having five valence electrons accept to satisfy thir octet. The semi conductors exist between 3rd and 5th groups exhibiting different properties according to the temperatures and excitations provided.

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If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much w

aleksklad [387]

Answer:3W

If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?

A) 2W

B) 3W

C) 4W

D) 6W

Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other is

W = k(+q)(+q)/ d

k is couloumb's constant

work done in moving 3 equal positive charges from infinity to a finite distance is given by

W₂=W₄=W₆=k(+q)(+q)/ d

Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d

= W + W + W = 3W

7 0

3 years ago

What is the example of stop moving

barxatty [35]

Answer:

Example:

A car applying brake.

A player catching a bowl.

8 0

3 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$