Phosphorus is implanted into silicon. The implant parameters are a dose of 1015cm-2at an energy of 150keV. a)Find the depth of the peak of the implant profile and its value at that depth. b)If the wafer originally had 1016cm-3of boron uniformly distributed throughout, find the depth(s) at which the concentration of phosphorus is equal to the concentration of boron.

a)  ≈ 4.99 / cm^3

b)   0.515 um

Explanation:

Given data:

energy = 150 Kev

dose of 10^15 cm^-2

A) find the depth of the peak of the implant profile and its value at that depth

expression for implant profile = written below in the attached file

value of the peak depth ≈ 4.99 / cm^3

B ) determine the depth at which the concentration of phosphorus is equal concentration of boron

= 0.515 um

attached below is the detailed solution to the problem

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.