The surface concentration of P and B needs to be calculated. Concept introduction: The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface. The expression for the Fick's second law is as follows: ∂ c ∂ t = ∂ ∂ x ( D ∂ c ∂ x ) After derivation: c s − c x c s − c 0 = erf ( x 2 D t ) .......... ( 1 ) Here, c s is the concentration of diffusion at the surface, c 0 is the concentration of diffusion at x, c 0 is the uniform concentration of diffused atoms in the material, c x is the concentration of atoms diffusing at x and D is diffusion coefficient.

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Answer 1

Question

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

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Answer 2

Question

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

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Answer 3

Question

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

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Answer 4

Question

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

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Answer 5

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

Answer 6

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.1×10−13cm2/s.

Substitute the above value in equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.1× 10 −13 )( 3600 ) )=erf(0.1)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.1, using table (1).

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

The value of surface concentration for both p type and n type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 hr will be 1.1×1018 atoms/cm3

Answer 7

Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of P and B needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth x under the surface.

The expression for the Fick's second law is as follows:

∂c∂t=∂∂x(D∂c∂x)

After derivation:

cs−cxcs−c0=erf(x2 Dt)..........(1)

Here, cs is the concentration of diffusion at the surface, c0 is the concentration of diffusion at x, c0 is the uniform concentration of diffused atoms in the material, cx is the concentration of atoms diffusing at x and D is diffusion coefficient.

Answer 8

Expert Solution & Answer

Answer to Problem 5.56P

Surface concentration for n type after 1 h, is 1.1×1018 atoms/cm3.

Surface concentration for p type after 1 h is 1.1×1018 atoms/cm3

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.

Time t when concentration is cx is 3600 s.

The diffusion coefficient D is 6.5×10−13cm2/s.

Substitute the above values in Fick's equation (1) as given below.

cs−cxcs−c0=erf(x 2 Dt )=erf( 1× 10 −5 2 ( 6.5× 10 −13 )( 3600 ) )=erf(0.103)

Find the value of (erfx2 Dt) when the value of (x2 Dt) is 0.103, using table (1).

Error function values for Fick's second law
Argument of the error x2Dt	Values of the error function erfx2Dt
0	0
0.10	0.1125
0.20	0.2227
0.30	0.3286
0.40	0.4284
0.50	0.5205
0.60	0.6039
0.70	0.6778
0.80	0.7421
0.90	0.7969
1.00	0.8427
1.50	0.9961
2.00	0.9953

erf(0.103)=0.1125cs−cxcs−c0=0.1125cs− 10 18cs−0=0.1125cs−1018=0.1125cscs= 10 180.8875cs=1.1×1018 atoms/cm3

Hence, Surface concentration for n type after 1 hr, will be 1.1×1018 atoms/cm3.

For the p type region

Here, the constant concentration of diffusing atoms at the surface is cs.

The uniform concentration of diffusing atoms in the material c0=0 (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location x=1×10−5cm is 1018 atoms/cm3.