What does the release or absorption of energy indicate?

Arsenic diffusion in Si: Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If the surf

Rainbow [258]

Answer:

Diffusion time is 7.42 h

Solution:

As per the question:

Temperature, T = $1100^{\circ}C$

Surface concentration of arsenic, $C_{S} = 5.0\times 10^{18}\ atoms/cm^{3}$

Surface concentration below Silicon surface, $C_{x} = 1.5\times 10^{16}\ atoms/cm^{3}$

D = $3.0\times 10^{- 14}\ cm^{2}/s$

x = $1.2\mu m = 1.2\times 10^{- 4}\ cm$

Initial concentration at t = 0, $C_{o} = 0$

Now, by using Flick's second eqn:

$\frac{C_{S} - C_{x}}{C_{x} - C_{o}} = erf(\frac{x}{\sqrt{Dt}})$

Thus by putting appropriate values:

$\frac{5.0\times 10^{18} - 1.5\times 10^{16}}{5.0\times 10^{18}} = erf(\frac{1.2\times 10^{- 4}}{2\sqrt{3.0\times 10^{- 14}t}})$

$0.997 = erf(\frac{364.4}{\sqrt{t}})$ (1)

Now,

$erf(z) = 0.997$

Now, from error function values tabulation:

For z = 2.0, erf(z) = 0.998

For z = 2.2, erf(z) = 0.995

Now,

With the help of linear interpolation method:

$\frac{z - 2}{2.2 - 2.0} = \frac{0.997 - 0.995}{0.998 - 0.995}$

z = 2.12

Now, using eqn (1) and above value:

$\frac{364.4}{\sqrt{t}} = 2.12$

$t = (\frac{364.4}{2.12})^{2}$ = 26700 s

t = $\frac{26700}{3600} = 7.42\ h$

7 0

3 years ago

When an electron absorbs energy it jumps to what state?

VMariaS [17]

Then, it jumps to HIGHER ORBITALS

4 0

4 years ago

What 3 dance styles make up musical theatre?

sveta [45]

Answer:

I believe its Jazz, Tao and Ballet, if im wrong then im sorry

Explanation:

3 0

3 years ago

Which of of the following is a physical change ice melting into a puddle of water,Leaves changing color in fall, Or baking a cak

Tom [10]

Answer:

ice melting

Explanation:

Because once it melts you can change it back to ice.

4 0

4 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

3 years ago