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Sedbober [7]
4 years ago
13

Find the number of free electrons per cubic centimeter in germanium at room temperature given that there are 5.6x10-10 free elec

trons per atom. The density and molar mass of silicon are 5.32 g/cm3 and 72.59 g/mol, respectively.
Physics
1 answer:
charle [14.2K]4 years ago
5 0

Answer:

2.41 \times 10^{13}\ free\ electrons/cm^3

Explanation:

given,

free electron per atom = 5.6 x 10⁻¹⁰

density of silicon = 5.32 g/cm³

molar mass of silicon =  72.59 g/mol

number of moles per cubic centimeter

           = \dfrac{5.32}{72.59}\ mole/cm^3

number of atom per cubic centimeter

           = \dfrac{5.32}{72.59}\times 6.022 \times 10^{23}\ /cm^3

number of free electron per cubic centimeter

           = \dfrac{5.32}{72.59}\times 6.022 \times 10^{23}\times 5.6 \times 10^{-10}/cm^3

           = 2.41 \times 10^{13}\ free\ electrons/cm^3

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Answer: The first answer for the first problem, and the 2nd answer for the second problem

Explanation: For the first one, if it is absolute zero, the molecules would not move at all.

For the second one, the temperature of the sample will increase due to the movement.

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When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s
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Explanation:

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3 years ago
Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w
amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

1)

  • Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
  • There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

        v = \lambda * f (1)

  • Replacing by the givens, and solving for f, we get:

       f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)

⇒     f = 8.6 Ghz (with two significative figures)

2)

  • Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

  • Solving for t, we get:

       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

       ⇒  t = 0.2 ms (with two significative figures)

6 0
3 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
4 years ago
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