318898
A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is \(408 \mathrm{pm}\). The diameter of the metal atom is
1 \(144 \mathrm{pm}\)
2 \(204 \mathrm{pm}\)
3 \(288 \mathrm{pm}\)
4 \(408 \mathrm{pm}\)
Explanation:
For \(f c c\) lattice \[ \begin{aligned} 4 r & =\sqrt{2 a} \\ r & =\dfrac{\sqrt{2}}{4} a=\dfrac{a}{2 \sqrt{2}}=\dfrac{408}{2 \sqrt{2}} \\ & =144 \mathrm{pm} \end{aligned} \] Diameter \((d)=2 r=2 \times 144 \mathrm{pm}=288 \mathrm{pm}\)
CHXII01:THE SOLID STATE
318899
If "a" stands for the edge length of the cubic systems: simple cubic, body-centred cubic, and face-centred cubic, then the ratio of radii of the spheres in these system will be respectively
\(s c, 2 r=a \quad \Rightarrow r=a / 2\) \(b c c, \quad 4 r=\sqrt{3} a \quad \Rightarrow r=\dfrac{\sqrt{3} \times a}{4}\) \(f c c, \quad 4 r=\sqrt{2} a \quad \Rightarrow r=\dfrac{a}{2 \sqrt{2}}\)
CHXII01:THE SOLID STATE
318900
Silver crystallises in fcc structure, if edge length of unit cell is \(316.5 \mathrm{pm}\). What is the radius of silver atom?
1 \(121.91 \mathrm{pm}\)
2 \(111.91 \mathrm{pm}\)
3 \(137.04 \mathrm{pm}\)
4 \(158.25 \mathrm{pm}\)
Explanation:
Given, Edge length of unit cell (a) \(=316.5 \mathrm{pm}\) For fcc we know, \[ \begin{aligned} \operatorname{Radius}(\mathrm{r}) & =\dfrac{\mathrm{a}}{2 \sqrt{2}} \\ & =\dfrac{316.5}{2 \sqrt{2}} \mathrm{pm} \\ = & \dfrac{158.25}{1.414} \mathrm{pm} \\ = & 111.91 \mathrm{pm} \end{aligned} \]
CHXII01:THE SOLID STATE
318901
Lithium forms body centred cube structure. The length of the side of its unit cell is \(351 \mathrm{pm}\). Atomic radius of the lithium will be :
318898
A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is \(408 \mathrm{pm}\). The diameter of the metal atom is
1 \(144 \mathrm{pm}\)
2 \(204 \mathrm{pm}\)
3 \(288 \mathrm{pm}\)
4 \(408 \mathrm{pm}\)
Explanation:
For \(f c c\) lattice \[ \begin{aligned} 4 r & =\sqrt{2 a} \\ r & =\dfrac{\sqrt{2}}{4} a=\dfrac{a}{2 \sqrt{2}}=\dfrac{408}{2 \sqrt{2}} \\ & =144 \mathrm{pm} \end{aligned} \] Diameter \((d)=2 r=2 \times 144 \mathrm{pm}=288 \mathrm{pm}\)
CHXII01:THE SOLID STATE
318899
If "a" stands for the edge length of the cubic systems: simple cubic, body-centred cubic, and face-centred cubic, then the ratio of radii of the spheres in these system will be respectively
\(s c, 2 r=a \quad \Rightarrow r=a / 2\) \(b c c, \quad 4 r=\sqrt{3} a \quad \Rightarrow r=\dfrac{\sqrt{3} \times a}{4}\) \(f c c, \quad 4 r=\sqrt{2} a \quad \Rightarrow r=\dfrac{a}{2 \sqrt{2}}\)
CHXII01:THE SOLID STATE
318900
Silver crystallises in fcc structure, if edge length of unit cell is \(316.5 \mathrm{pm}\). What is the radius of silver atom?
1 \(121.91 \mathrm{pm}\)
2 \(111.91 \mathrm{pm}\)
3 \(137.04 \mathrm{pm}\)
4 \(158.25 \mathrm{pm}\)
Explanation:
Given, Edge length of unit cell (a) \(=316.5 \mathrm{pm}\) For fcc we know, \[ \begin{aligned} \operatorname{Radius}(\mathrm{r}) & =\dfrac{\mathrm{a}}{2 \sqrt{2}} \\ & =\dfrac{316.5}{2 \sqrt{2}} \mathrm{pm} \\ = & \dfrac{158.25}{1.414} \mathrm{pm} \\ = & 111.91 \mathrm{pm} \end{aligned} \]
CHXII01:THE SOLID STATE
318901
Lithium forms body centred cube structure. The length of the side of its unit cell is \(351 \mathrm{pm}\). Atomic radius of the lithium will be :
318898
A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is \(408 \mathrm{pm}\). The diameter of the metal atom is
1 \(144 \mathrm{pm}\)
2 \(204 \mathrm{pm}\)
3 \(288 \mathrm{pm}\)
4 \(408 \mathrm{pm}\)
Explanation:
For \(f c c\) lattice \[ \begin{aligned} 4 r & =\sqrt{2 a} \\ r & =\dfrac{\sqrt{2}}{4} a=\dfrac{a}{2 \sqrt{2}}=\dfrac{408}{2 \sqrt{2}} \\ & =144 \mathrm{pm} \end{aligned} \] Diameter \((d)=2 r=2 \times 144 \mathrm{pm}=288 \mathrm{pm}\)
CHXII01:THE SOLID STATE
318899
If "a" stands for the edge length of the cubic systems: simple cubic, body-centred cubic, and face-centred cubic, then the ratio of radii of the spheres in these system will be respectively
\(s c, 2 r=a \quad \Rightarrow r=a / 2\) \(b c c, \quad 4 r=\sqrt{3} a \quad \Rightarrow r=\dfrac{\sqrt{3} \times a}{4}\) \(f c c, \quad 4 r=\sqrt{2} a \quad \Rightarrow r=\dfrac{a}{2 \sqrt{2}}\)
CHXII01:THE SOLID STATE
318900
Silver crystallises in fcc structure, if edge length of unit cell is \(316.5 \mathrm{pm}\). What is the radius of silver atom?
1 \(121.91 \mathrm{pm}\)
2 \(111.91 \mathrm{pm}\)
3 \(137.04 \mathrm{pm}\)
4 \(158.25 \mathrm{pm}\)
Explanation:
Given, Edge length of unit cell (a) \(=316.5 \mathrm{pm}\) For fcc we know, \[ \begin{aligned} \operatorname{Radius}(\mathrm{r}) & =\dfrac{\mathrm{a}}{2 \sqrt{2}} \\ & =\dfrac{316.5}{2 \sqrt{2}} \mathrm{pm} \\ = & \dfrac{158.25}{1.414} \mathrm{pm} \\ = & 111.91 \mathrm{pm} \end{aligned} \]
CHXII01:THE SOLID STATE
318901
Lithium forms body centred cube structure. The length of the side of its unit cell is \(351 \mathrm{pm}\). Atomic radius of the lithium will be :
318898
A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is \(408 \mathrm{pm}\). The diameter of the metal atom is
1 \(144 \mathrm{pm}\)
2 \(204 \mathrm{pm}\)
3 \(288 \mathrm{pm}\)
4 \(408 \mathrm{pm}\)
Explanation:
For \(f c c\) lattice \[ \begin{aligned} 4 r & =\sqrt{2 a} \\ r & =\dfrac{\sqrt{2}}{4} a=\dfrac{a}{2 \sqrt{2}}=\dfrac{408}{2 \sqrt{2}} \\ & =144 \mathrm{pm} \end{aligned} \] Diameter \((d)=2 r=2 \times 144 \mathrm{pm}=288 \mathrm{pm}\)
CHXII01:THE SOLID STATE
318899
If "a" stands for the edge length of the cubic systems: simple cubic, body-centred cubic, and face-centred cubic, then the ratio of radii of the spheres in these system will be respectively
\(s c, 2 r=a \quad \Rightarrow r=a / 2\) \(b c c, \quad 4 r=\sqrt{3} a \quad \Rightarrow r=\dfrac{\sqrt{3} \times a}{4}\) \(f c c, \quad 4 r=\sqrt{2} a \quad \Rightarrow r=\dfrac{a}{2 \sqrt{2}}\)
CHXII01:THE SOLID STATE
318900
Silver crystallises in fcc structure, if edge length of unit cell is \(316.5 \mathrm{pm}\). What is the radius of silver atom?
1 \(121.91 \mathrm{pm}\)
2 \(111.91 \mathrm{pm}\)
3 \(137.04 \mathrm{pm}\)
4 \(158.25 \mathrm{pm}\)
Explanation:
Given, Edge length of unit cell (a) \(=316.5 \mathrm{pm}\) For fcc we know, \[ \begin{aligned} \operatorname{Radius}(\mathrm{r}) & =\dfrac{\mathrm{a}}{2 \sqrt{2}} \\ & =\dfrac{316.5}{2 \sqrt{2}} \mathrm{pm} \\ = & \dfrac{158.25}{1.414} \mathrm{pm} \\ = & 111.91 \mathrm{pm} \end{aligned} \]
CHXII01:THE SOLID STATE
318901
Lithium forms body centred cube structure. The length of the side of its unit cell is \(351 \mathrm{pm}\). Atomic radius of the lithium will be :
