Chemical analysis of a silicon crystal reveals gallium at a level of 3.091×10−8 atomic percent. Assuming that the concentration of thermally excited charge carriers from the Si matrix is negligible, what is the density of free charge carriers (free carriers/cm3) in this Si crystal?
The answer 1.598e14 was not right for me too. Any ideas what happen and how to write it out?
As its Si: 5.00*10^22 / (gallium level*10^-2)
Thank you someguy
how can you do this one?
What is the formula?, please
but answer is not right its not 5.00*10^22
guys check ur index value of the gallium levels they are different
CHEAT!
These are midterm exam problems. They shouldn't be posted here. Anyone who does Deserves yo be kicked off the course.
(6.023*10^23)/(12.05)=m
m*(3.091*10^-8)*(10^-2)=
To determine the density of free charge carriers in the silicon crystal, we need to know the number of gallium atoms present and the volume of the crystal.
The atomic percent of gallium in the crystal is given as 3.091×10−8 atomic percent. This means that for every 100 million silicon atoms, there are 3.091 gallium atoms.
Now, we need to convert the atomic percent of gallium to the actual number of gallium atoms in the crystal. To do this, we need to know the total number of silicon atoms in the crystal. Let's assume this number is N_si.
Using the given atomic percent, we can set up the following equation:
(3.091 × 10^(-8) / 100) × N_si = number of gallium atoms (N_ga)
Next, we need to calculate the number of free charge carriers (N_fc) in the crystal. In an intrinsic semiconductor like silicon, each dopant atom (gallium, in this case) contributes one free charge carrier. So N_fc = N_ga.
Finally, we need to calculate the density of free charge carriers (ρ_fc) in terms of carriers per cm^3. We can divide the number of free charge carriers by the volume (V) of the crystal:
ρ_fc = N_fc / V
Please specify the volume of the silicon crystal to proceed with the calculation.