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Alenkasestr [34]
2 years ago
13

The resistivity of a certain semi-metal is 10-3 Ohm-cm. Suppose we would like to prepare a silicon wafer with the same resistivi

ty. Assuming we will use n-type dopants only, what dopant density would we choose

Physics
1 answer:
faltersainse [42]2 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

  The dopant density is ND   ≈  8.135*10¹²  cm⁻³

Explanation:

The explanation is shown on the second , third ,fourth and fifth uploaded image

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Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are
Juliette [100K]

Answer:Q=248.011 W

Explanation:

Given

Temperature of Room T_{\infty }=18^{\circ}\approx 291 K

Area of Person A=1.7 m^2

Temperature of skin T=32^{\circ}\approx 305 K

Heat transfer coefficient h=5 W/m^2.k

Emissivity of the skin and clothes \epsilon =0.9

\Delta T=32-18=14^{\circ}

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )

where \sigma =stefan-boltzman\ constant

Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)

Q_1=129.01 W

Heat Transfer due to convection is given by

Q_2=hA(\Delta T)

Q_2=5\times 1.7\times 14=119 W

Q=Q_1+Q_2

Q=129.01+119=248.011 W

7 0
3 years ago
The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation
makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

3 0
3 years ago
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Base
shepuryov [24]

Answer:

Explanation:

 For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C)  = μ

sin65 / cos C = μ

sinC = 1.33 / μ  , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6   Ans

3 0
2 years ago
Explain how the fixed points are used when calibrating a thermometer. ​
8090 [49]

<u>Answer:</u>

First, the thermometer is dipped into boiling water, and the mercury inside the thermometer rises to a high level, called the boiling point. This level is then marked as 100°C. The thermometer is then dipped into melting ice, which causes the mercury level to fall to a point called the ice point. This point is then marked as 0°C. The length of the thermometer from the 0°C mark to the 100°C point is then divided into 100 equal sections, and the rest of the levels are marked accordingly.

8 0
2 years ago
A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.
scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

                                           = \sqrt{[37.084]^{2}+[44.984]^{2}  }

                                           =\sqrt{3398.78}

                                     OB =58.3

                                         

7 0
9 months ago
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