Question

14
At
23 k the crystal structure
transforms From BCC to HCP
keeping density The cubic face
B ce is o2= 3.42 A² and in
HCP is c = 1.53 The lattic
|
a
constant
in
HCP
structure​

Answer 1

Answer:

For BCC

$4r = \sqrt{3a}$

$r = \frac{ \sqrt{3 \times 3.42} }{4}$

$= 1.48$

For HCP

$a = 2r$

$= 2 \times 1.48$

$a = 2.96 \: {a}^{2}$