Tomic radius INCREASES as you go DOWN a Group because each successive Period (row) has an additional occupied energy level. you can think of it this way: each time you drop down a row, you add a "ring."

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For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

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