Answer
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Hint: We are given density, atomic mass, Avogadro number of the lithium metal. So to find the edge length of a unit cell of lithium metal, we have to link all the given entities. So we can calculate density in terms of atomic mass, avogadro number and edge length as given in the below formula. Use it and find the value of the required entity.
Complete Step by step answer: We are given that Lithium has a bcc structure. Its density is $530kg{m^{ - 3}}$ and its atomic mass is $6.94gmo{l^{ - 1}}$.
We have to find the edge length of its unit cell.
Density d of a metal is $d = \dfrac{{ZM}}{{{a^3}{N_A}}}$. We have to use this formula to find the edge length.
We are given that lithium has a bcc structure.
A bcc unit cell totally consists of two atoms which is our Z.
Density is given in Kilometres per cubic meters and Atomic mass is given in gram per moles. So convert density into gram per cubic centimeters.
$
d = 530kg/{m^3} \\
\Rightarrow d = 530 \times 0.001g/c{m^3} = 0.530g/c{m^3} \\
Z = 2,M = 6.94g/mol,{N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}} \\
\Rightarrow d = \dfrac{{ZM}}{{{a^3}{N_A}}} \\
\Rightarrow 0.53 = \dfrac{{2 \times 6.94}}{{{a^3} \times 6.02 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = \dfrac{{2 \times 6.94}}{{0.53 \times 6.02 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = \dfrac{{13.88}}{{3.1906 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = 4.35 \times {10^{ - 23}} \\
\therefore a = 3.516 \times {10^{ - 8}}cm \approx 352 \times {10^{ - 12}}m = 352pm \\
$
Therefore, the edge length of a unit cell of Lithium metal is 352 pm.
Hence, the correct option is Option B.
Note: BCC (body-centered cubic) unit cells have atoms at each of the eight corners of a cube and one atom at the center. Each corner atom will be shared among eight unit cells. Do not confuse bcc structure with fcc structure which is face-centered cubic. In fcc, atoms are arranged at the corners of the cell and at the center of each cube face. The difference is bcc has overall 2 atoms whereas fcc has 4 atoms. So be careful while mentioning the no. of atoms.
Complete Step by step answer: We are given that Lithium has a bcc structure. Its density is $530kg{m^{ - 3}}$ and its atomic mass is $6.94gmo{l^{ - 1}}$.
We have to find the edge length of its unit cell.
Density d of a metal is $d = \dfrac{{ZM}}{{{a^3}{N_A}}}$. We have to use this formula to find the edge length.
We are given that lithium has a bcc structure.
A bcc unit cell totally consists of two atoms which is our Z.
Density is given in Kilometres per cubic meters and Atomic mass is given in gram per moles. So convert density into gram per cubic centimeters.
$
d = 530kg/{m^3} \\
\Rightarrow d = 530 \times 0.001g/c{m^3} = 0.530g/c{m^3} \\
Z = 2,M = 6.94g/mol,{N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}} \\
\Rightarrow d = \dfrac{{ZM}}{{{a^3}{N_A}}} \\
\Rightarrow 0.53 = \dfrac{{2 \times 6.94}}{{{a^3} \times 6.02 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = \dfrac{{2 \times 6.94}}{{0.53 \times 6.02 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = \dfrac{{13.88}}{{3.1906 \times {{10}^{23}}}} \\
\Rightarrow {a^3} = 4.35 \times {10^{ - 23}} \\
\therefore a = 3.516 \times {10^{ - 8}}cm \approx 352 \times {10^{ - 12}}m = 352pm \\
$
Therefore, the edge length of a unit cell of Lithium metal is 352 pm.
Hence, the correct option is Option B.
Note: BCC (body-centered cubic) unit cells have atoms at each of the eight corners of a cube and one atom at the center. Each corner atom will be shared among eight unit cells. Do not confuse bcc structure with fcc structure which is face-centered cubic. In fcc, atoms are arranged at the corners of the cell and at the center of each cube face. The difference is bcc has overall 2 atoms whereas fcc has 4 atoms. So be careful while mentioning the no. of atoms.
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