Question

Barium metal crystallizes in a body centered cubic lattice. the unit cell edge length is 502 pm. calculate the atomic radius of ba.

Answer 1

Answer:

The atomic radius of ba is $= 2.17 \times 10^{-10}m$  or 217. 37 pm

Explanation:

For a body centred cubic lattice,  the atomic radius is calculated by using the formula

$r = \frac{\sqrt{3} \times a}{4}$

where a = unit cell edge length = 502 pm

Next, to apply the formula properly, we will have to convert everything to S.I units.

To do this, we will divide 502 picometers by $1 \times 10^{12} m$ = $502 \times 10^{-12} m$

$r = \frac{\sqrt{3} \times 502 \times 10^{-12}}{4}= 2.17 \times 10^{-10}m$

we can convert this back to pm by multiplying by $1\times 10^{12}$

Hence, r = 217. 37 pm

Answer 2

So, the radius of $ba$ an atom is $217.4 pm$.

Atomic radius:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

The unit cell edge length is 502 pm that is $502\times 10^{-10} cm$

The radius of a Barium atom in a cubic lattice is related to the unit cell edge length is as follow,

$r=\frac{a\sqrt{3} }{4}$

Where $r$ is the radius and $a$ is the unit cell length

$r=502\times 10^{-10}\times \frac{\sqrt{3} }{4}\\=2174 \times 10^{-3}\\r=217.4 pm$

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