A hypothetical metal has a tetragonal unit cell for which the lattice parameters (in nm) are a=b=0.550, c=0.673 and all angles are 90 degrees. If there are 5 atoms per unit cell and the atomic packing factor is 0.63, determine the atomic radius in nm. Use 3.14 for pi. Give your answer to 3 decimal places. I got 0.197 the first time and it was incorrect.
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A hypothetical metal has a tetragonal unit cell for which the lattice parameters (in nm) are a=b=0.550, c=0.673 and all angles are 90 degrees. If there are 5 atoms per unit cell and the atomic packing factor is 0.63, determine the atomic radius in nm. Use 3.14 for pi. Give your answer to 3 decimal places.
I got 0.197 the first time and it was incorrect.
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- At room temperature, the edge length of the cubic unit cell in elemental silicon is 5.431 Å, and the density of silicon at the same temperature is 2.328 g cm-3 . Each cubic unit cell contains eight silicon atoms. Using only these facts, perform the following operations.(a) Calculate the volume (in cubic centimeters) of one unit cell.(b) Calculate the mass (in grams) of silicon present in a unit cell.(c) Calculate the mass (in grams) of an atom of silicon.(d) The mass of an atom of silicon is 28.0855 u. Estimate Avogadro’s number to four significant figures.The accompanying figure shows three different crystallographic planes for a unit cell of a hypothetical metal. The circles represent atoms. Draw its unit cell and show the atoms’ positions. To what crystal system does the unit cell belong and what would this crystal structure be called?Calculate the volume of a unit cell of Nickel in nm3. The volume of unit cell of Nickel is 4.3142×10-29 m3. Note that Nickel has a Face Centred Cubic Lattice Structure. The atomic mass of nickel is 58.78 g/mole. Find the density of Nickel in Mg/m3. (Here, Mg is defined as Mega-gram) Use the value of Avogedro Number NA=6.023×1023 Use the below mentioned formulae to calculate the density (ρ) ρ=(n×A)/(NA×Vc ), Here, A = the atomic mass of Nickel, Vc=Volume of unit cell of nickel, n=number of atoms in an unit cell of Nickel.
- A hypothetical metal with an atomic radius of 0.171 nm has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, and 0.725, and 0.978 nm, respectively. It has an atomic weight of 126.91 g/mol and a theoretical density of 4.92 g/cm3. Determine the APF of the metal. Use NA = 6.022 x 1023Which of the following statements is/are CORRECT? If an ionic compound with the formula MX forms a face-centered cubic unit cell with the anions (Xn–) at the lattice points, the cations (Mn+) may occupy 1. one fourth of the tetrahedral holes in each unit cell. 2. all of the octahedral holes in each unit cell. 3. the center of each face in each unit cell. (a) 1 only (b)2 only (c)3 only (d)1 and 2 (e)1, 2, and 31) Calculate the percentage of empty space in a simple, body-centered, and face-centered cubic unit cell using geometry and algebra without substituting any values for atomic radius (r) or cell length (a). The volume of a sphere is 4πr3 3 Hint: Express "a" as a function of "r." The variable "r" will cancel out when you calculate the volume of "atoms" in the cell to give a number which is the fraction of occupied space. The rest of the space is empty.
- 1.Indium has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. (a) If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of indium is 114.82 g/mol; compute its theoretical density.Determine the empirical formula for the following compound. Use the notation W, Yy. Assume that W is a cation and Y is an anion. Determine the coordination number for the yellow spheres (i.e., how many white ions each yellow ion is adjacent to). Determine the coordination number for the white spheres (smaller spheres like those on the corners). Empirical Formula W6Y8 Coordination number for yellow spheres 6 Coordination number for white spheres 8The density of an unknown metal is 2.64 g/cm3 and its atomic radius is 0.215 nm. It has a face-centered cubic lattice. Determine its atomic weight.
- 1. Simple Cubic Unit Cell a. Consider one unit cell and assume the length of the side of the cube is "a". Remember that “a" is the distance between the centers of two adjacent atoms. How long is “a", the edge of a unit cell, in terms of radius, r, of an atom? b. A simple cubic cell has the equivalent of only 1 atom. Recall the volume of sphere with radius, r, is expressed as V = 4/3 r. With this information, find the total volume of all the spheres in this unit cell, expressed in terms of r. To do this, take the total number of atoms and multiply it by the volume of one atom, with radius, r) c. Find the volume of the entire unit cell including the empty space in terms of r. Remember for a cube the volume, V=1 xw xh. You already know the edge length of the side in terms of "r" based in the earlier questions. d. Find the fraction of the total volume of the cube (which is the unit cell's volume) is occupied by the atoms. Express this as percent and write your answer in the summary table.number 3 onlyMany elements crystalize with a unit cell that is hexagonally shaped (6 sided). The calculations of the unit cell dimension are not significantly more complicated than those for cubic unit cells. A hexagonal unit cell is shown below.The cell dimension "a" is twice the radius (r) of the atom. 2r=a. For most elements the dimension "c" is approximately 1.633×"a".The element technetium (Tc) crystallizes with a Simple Hexagonal unit cell. The density of a technetium is 11.5 g/cm3. Use this information to calculate the metallic radius of technetium in picometers (pm). 1 pm = 1×10−12 meters. You may assume that the dimension c=1.633·a