Question

A metal M has body centred cubic crystal. The edge length of the unit cell of the crystal is 0.602 nm. Determine the radius of M atom and volume occupied by the atoms in the unit cell.​

Answer 1

Answer:

Solution

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Correct option is A)

The expression for density is given below:

Z=2 for BCC, M= 50.3 g/ mol, a=0.304 nm=0.304×10

−7

cm and N

A

=6.0×10

23

d=

N

A

a

3

zM

=

6×10

23

×(0.304×10

−7

)

3

2×50.3

=5.945 g cm

−3

Option A is correct.

Video Explanation

Answer 2

Answer:

The radius of the atom in the metal M which crystallizes with the body-centered unit cell is  $2.606$ × $10^{-10}$ m and the volume occupied by the atoms in that unit cell is $1.482\ *\ 10^{-28}\ m^3$·

Explanation:

Given that,

The metal M crystallizes with body-centered unit cell·

The edge length of the unit cell, a  $=$  $0.602\ nm\ =\ 0.602$ × $10^{-9} \ m$

To find out the radius of the atoms in M metal,

The relation that connects the radius with the edge length for the bcc is,

    $\sqrt{3}\ a\ =\ 4\ r$

where,

a $=$ edge length of the unit cell

r $=$ radius of the atoms in the metal M

    $\sqrt{3}\ a\ =\ 4\ r$

⇒  $r\ =\ \frac{\sqrt{3}\ a }{4}$

On substituting the value we get,

⇒  $r\ =\ \frac{\sqrt{3}\ *\ 0.602\ *\ 10^{-9}\ m }{4}$

⇒  $r\ =\ 2.606$ × $10^{-10}$ m

Therefore, the radius of the atom in the metal M is  $2.606$ × $10^{-10}$ m·

Now,

To find out the volume occupied by the atoms in the unit cell,

We know that,

Each unit cell in the bcc structure contains $2$ atoms and the atoms are assumed to be a spherical shape·

Therefore,

The volume occupied by the atoms in the unit cell V is,

 V $=$ $2$ × $\frac{4}{3}\ \pi \ r^{3}$

where,

r $=$ radius of the atoms in the metal M  $=$  $2.606$ × $10^{-10}$ m·

π $=$ $3.14$

On substituting the values we get,

⇒  $V =\ 2\ *\ \frac{4}{3}\ *\ 3.14\ *\ (2.606\ *\ 10^{-10})^{3}$

⇒  $V\ =\ 1.482\ *\ 10^{-28}\ m^3$

Hence, the volume occupied by the atoms in the unit cell is $1.482\ *\ 10^{-28}\ m^3$·