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In a body-centered cubic unit cell, there are two atoms per unit cell. Hence, from the expression r=N/a^3(M/Na)
We get, a^^3=N/ρ(M/Na)
=(2/10.3gcm^−3)x(95.04gmol^−1 /6.023×1023mol^−1)
=3.09×10^−23cm^3 or a=3.13×10^−8cm=313pm
Now since in the body centered cubilc unit cell , atoms touch each other along the cross diagonal of the cube , we have 4r=√3a
r=(√3a)(313pm)/4=135.5pm
so ans would be (3) 135.96
I have taken real value so approx ans...
PLS MAKE IT THE BRAINLIEST
In a body-centered cubic unit cell, there are two atoms per unit cell. Hence, from the expression r=N/a^3(M/Na)
We get, a^^3=N/ρ(M/Na)
=(2/10.3gcm^−3)x(95.04gmol^−1 /6.023×1023mol^−1)
=3.09×10^−23cm^3 or a=3.13×10^−8cm=313pm
Now since in the body centered cubilc unit cell , atoms touch each other along the cross diagonal of the cube , we have 4r=√3a
r=(√3a)(313pm)/4=135.5pm
so ans would be (3) 135.96
I have taken real value so approx ans...
PLS MAKE IT THE BRAINLIEST
mrigansh:
PLS MAKE IT THE BRAINLIEST
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