Chemistry, asked by shauryashivani4657, 1 year ago

What is the radius of sodium atom if it crystallises in bcc structure with the cell edge of 400 pm?

Answers

Answered by matiullahansarp5rluk
3
{(400*10^-3)*√3)÷4} nm
Answered by skanthah
1

Answer:

300 pm

Explanation:

For bcc, r=(\sqrt{3}/4) * 400

r=100\sqrt{3}

r=100*1.732

r=173.2 pm

Similar questions