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1 year ago

Find the electron and hole concentrations and fermi level in silicon at 300 k for boron

1 answer:

4 0

To find the electron and hole concentrations and fermi levels in silicon at 300 k for boron, fermi level is 9.3 x $10^4 cm^-3$ .

<h3>Calculation and Explanation</h3>

Ionization energy for boron in Si, 0.045 eV

Fermi level in silicon,

300 K at $10^{15} cm^{-3}$ (Boron atoms)

All boron impurities are ionized, 300K

Electron, hole concentrations and Fermi levels is found out,

The value of Na = $10^{15} cm^{-3}$

np = $\frac{ni^2}{nA}$

np = $\frac{(9.65 x 10^9)^2}{10^15}$

np= 9.3 x $10^4 cm^-3$

Fermi level is calculated,

Ef - Ev = kT ln(NV/ND)

value of kT = 0.0259 eV (300° K)

So, substitution the values

Ef - Ev = kT ln(NV/ND)

Ef - Ev = 0.0259ln (2.66 x $10^{19} / 10^{15}$ )

Ef - Ev = 0.0259ln (26600)

Ef - Ev = 0.0259 x 10.18

Ef - Ev = 0.263 eV

What is ionization energy?

Ionization energy, also known as ionization energy (IE) or ionization energy (British English spelling), is the minimal amount of energy needed to free the most loosely bonded electron from an isolated gaseous atom, positive ion, or molecule in physics and chemistry.

To know more about ionization energy, visit:

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A 300N box on a 43 degree angle.

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Answer:is this a question??? I’m so confused

Explanation:

7 0

3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

3 years ago

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep

Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = $\frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}$

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 - 0:00 = 5

D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 - 0:00 = 6

D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 - 0:00 = 12

D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0

3 years ago

A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) = 1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0

3 years ago

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)

Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

$W=mg\Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the object in this problem, we have:

m = 5 kg

$g=10 m/s^2$

$\Delta h = 2 m$

Substituting into the equation,

$W=(5)(10)(2)=100 J$

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3 0

3 years ago