Germanium has a cubic unit cell with a side edge of 565pm. The density of germanium is 5.36g/cm^3. What is the crystal system adopted by germaium.
Find the mass in grams in one cell.
mass=volume*denstiy
volume=(565pm)^3=1.8E-22cm^3
mass=1.8E-22 cm^3*5.36g/cm^3= you do it.
Ok, how many Ge atoms is that?
Now, given the number of atoms per unit cell, is it face centered, body centered, or simple cubic
I get 8 atom/unit cell and the answer in my book was face centered. But i am a bit confused don't FCC have 4 atoms/unit cell?
Yes, fcc cells have four atoms/unit cell; however, if y ou look at the link provided by Bob Pursley, you will see two things there. First, the statement that Ge is like diamond and the Ge structure is "two interpenetrating face centered unit cells". I was confused, also, when I worked the problem and obtained 8. I IMed Bob P, we discussed it, and he found the link shown above. In fact, we talked about how the student was supposed to know that Ge had the diamond structure and that it was interpenetrating face centered unit cells? We wondered if this was an upper upper class or a grad student problem. Both of us thought the problem was too difficult for lower level classes but of course we aren't teaching the course. I hope this helps explain.
Well, it sounds like Germanium is quite the trendsetter with its cubic unit cell! With a side edge of 565 pm and a density of 5.36 g/cm^3, Germanium is rocking the cubic crystal system. It just loves being all square and balanced.
To determine the crystal system adopted by germanium, we need to analyze the side length of its unit cell.
The cubic unit cell is characterized by having all of its edges of equal length. In this case, the provided side edge length is 565 pm (picometers), which is equivalent to 565 x 10^(-12) meters.
To determine the crystal system, we first need to calculate the volume of the unit cell. The volume (V) of a cubic unit cell can be calculated using the formula:
V = a^3
where "a" represents the side length of the unit cell.
Substituting the given side edge of 565 pm (or 565 x 10^(-12) meters) into the formula, we have:
V = (565 x 10^(-12))^3 m^3
Next, we need to convert the density from grams per cubic centimeter (g/cm^3) to kilograms per cubic meter (kg/m^3). We can use the conversion factor:
1 g/cm^3 = 1000 kg/m^3
Now, we can calculate the density in kg/m^3:
Density = 5.36 g/cm^3 x (1000 kg/m^3 / 1 g/cm^3)
By multiplying these values, we get:
Density = 5360 kg/m^3
Now, we have both the volume (V) and the density (Density) of germanium. We can use these values to determine the crystal system by comparing the density with known values for different crystal systems.
For germanium, the density falls within the range of 5000-6000 kg/m^3, which is characteristic of the diamond crystal system.
In conclusion, the crystal system adopted by germanium is the diamond crystal system.