Question

For a 100 keV boron implant with a dose of $5 \times 10^{14} \text{cm}^{-2}$, calculate peak concentration. Solution: From Fig. 6a, $R_p = 0.31 \text{ µm}$ and $\sigma_p = 0.07 \text{ µm}$

          For a 100 keV boron implant with a dose of $5 \times 10^{14} \text{cm}^{-2}$,
calculate peak concentration.
Solution:
From Fig. 6a, $R_p = 0.31 \text{ µm}$ and $\sigma_p = 0.07 \text{ µm}$
        
For a 100 keV boron implant with a dose of 5 × 10^14cm^-2,
calculate peak concentration.
Solution:
From Fig. 6a, Rp = 0.31  µm and = 0.07  µm

Added by Sandra J.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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For a 100 keV boron implant with a dose of 5 1014 cm2 calculate peak concentration Solution; From Fig.6a,R=0.31m and =0.07m
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Transcript

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00:01 Now for question number a, docs h is equals to beam current divided by q times implant time divided by beam area.
00:27 Now putting in the values we will have 1 times 10 to the power minus 3 divided by charge as 1 .6 times 10 to the power minus 9 coulomb times 5 seconds divided by 10 to the power 4 millimeter square.
00:45 This will give us the value 3 .125 times 10 to the power 12 millimeter to the power minus 2 or 3 .125 times 10 to the power 14 centimeter inverse square.
01:02 Now for question number b, peak concentration np we will have r is equals to root over 2 pi np delta rp.
01:22 Therefore np will be equals to r divided by root over 2 pi delta rp...
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