1.13   Niobium crystallises in body-centred cubic structure. If density is 8.55\; g\; cm^{-3}, calculate atomic radius of niobium using its atomic mass 93 u.

Answers (1)

Here we will use the relation between density and edge length.

                                          Density = \frac{z.M}{a^3.N_A}

It is given that Niobium crystallises in body-centred cubic structure so the value of z = 2.

So we will put the value of density, molar mass and z.

We get,                            

                                          8.55 = \frac{2\times93}{a^3\times6.022 \times10^{23}}

or                                       a^3 = 36.124 \times10^{-24}

or                                       a = 3.3057 \times10^{-8}\ cm

We know the relation between the radius of the atom and edge length in bcc lattice.

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

we get the radius of niobium atom    = 143.1 pm.

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