Lithium has a bcc structure. Its densit"
530 kg m-3 and its atomic mass is 6.94 gm
Calculate the edge length of a unit cell of Lith
metal (NA = 6.02 x 1023 mol-1)
(1) 264 pm (2) 154 pm
(3) 352 pm (4) 527 pm
Answers
Answer:
As we learnt in
Density of cubic unit cell -
d = \frac{zM}{a^3N_o}
- wherein
Where,
d = density of crystal
z = no. of effective constituent particles in one unit cell
M = molecular weight
a = edge length of unit cell
No = 6.022*1023
Density of cubic unit cell
d=\frac{zM}{a^{3}N_{o}}
a=\left ( \frac{zM}{dN_{o}} \right )^{\frac{1}{3}}
=\left ( \frac{2\times 6.94}{0.530g\times 6.02\times 10^{23}} \right )^{\frac{1}{3}}
=3.52\times 10^{-10}\: m
=352\: pm
Option 1)
154 pm
Option is incorrect
Option is incorrectOption 2)
352 pm
Option is correct
Option 3)
527 pm
Option is incorrect
Option 4)
264 pm
Option is incorrect
I hope it will help u
Answer:
Explanation:
As we learnt in
Density of cubic unit cell -
d = \frac{zM}{a^3N_o}
- wherein
Where,
d = density of crystal
z = no. of effective constituent particles in one unit cell
M = molecular weight
a = edge length of unit cell
No = 6.022*1023
Density of cubic unit cell
d=\frac{zM}{a^{3}N_{o}}
a=\left ( \frac{zM}{dN_{o}} \right )^{\frac{1}{3}}
=\left ( \frac{2\times 6.94}{0.530g\times 6.02\times 10^{23}} \right )^{\frac{1}{3}}
=3.52\times 10^{-10}\: m
=352\: pm
Option 1)
154 pm
Option is incorrect
Option 2)
352 pm
Option is correct
Option 3)
527 pm
Option is incorrect
Option 4)
264 pm
Option is incorrect