# Niobium has a density of 8.57 g/cm3 and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom

Here is my question

Niobium has a density of $8.57 \pu{g/cm^3}$ and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom.

Here is my solution • The thing I cannot get is there are too many different answer for this atomic radius questions. I mean like I've seen three different answer on websites. I've rechecked my question but couldn't find anymistake. Take a look here: gordonengland.co.uk/elements/nb.htm and here periodni.com/nb.html Here is anohter different answer too: elementsdatabase.com/Niobium-Nb-41-element Are those all radius correct? – Mascular Dec 4 '17 at 19:32
• That answer is clearly wrong because you do not give units! Please do so, or else we may think that your answer is $100$ times as large as you likely intended (assuming length is in S.I. units). – Oscar Lanzi Dec 4 '17 at 19:56
• The answer is not certain. As I said take a look that sites. You will see how right I am. Everyone finds different answers. According to Wikipedia, it seems $1.46 \text {pm}$. – Mascular Dec 4 '17 at 20:22
• According to Wikipedia, answer is $1.46$, on other sites, $1.45$ and $1.44$. on my textbook, $1.43$. Moreover, I rechecked my solutions twice. I'm sure about I didn't do mistake. – Mascular Dec 4 '17 at 20:33
• Uh ... how is the radius of an atom roughly $1$ picometer? I seem to recall that the Bohr radius of the hydrogen atom is $51$ picometers and that is just hydrogen. I would guess that the decimal points in the answers quoted here are not really there! – Oscar Lanzi Dec 4 '17 at 23:49

### This is how I would do the calculation...

Given data:

$$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$ $$\text {Density} = \pu{8.57 g/cm^3}$$

Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within the multiple calculations. I'll round the final result to 3 significant figures.

$$\text{Volume of mole} =\text{V}_{mole} = \frac {92.906}{8.57} = \pu{10.841 cm^3/mol}$$

In BCC, there are 2 atoms in one unit cell. Also note that the accepted figure for Avagadro's constant to five significant figures is now $6.0221\times10^{23}$

$$\text{Volume of unit cell} = \text{V}_{cell} = 10.841 \times \frac {2}{6.0221 \times 10^{23}} = 3.6004 \times 10^{-23}\text{ cm}^3$$ A unit cell is a cube with each side being $a$ $$\text{V}_{cell} = a^3$$ $$\therefore a = \sqrt{3.6004 \times 10^{-23}} = 3.3021\times 10^{-8}\text{ cm}$$

But the BCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = \dfrac{4}{\sqrt{3}}r$$ so

$$r = \dfrac{\sqrt{3}\times(3.3021\times 10^{-8})}{4} = 1.4299\times 10^{-8}\text{ cm}$$

Now rounding $r$ to 3 significant figures gives

$$r = 1.43\times 10^{-8}\text{ cm} = 143\text{ pm}$$

You shouldn't chase values through websites and literature. You can only calculate given whatever data you're suppose to take as given. For instance no temperature was given for the density which certainly changes with temperature. Also, to make the point again, with a calculator you should carry extra significant figures in the intermediate calculations to try to avoid round off errors in the final result(s).