Flag IIT JEE Entrance Exam> iit jee 2008
question mark

height of hcp unit cell

mukal meherchandani , 12 Years ago
Grade 12
anser 1 Answers
Aman Bansal

Last Activity: 12 Years ago

Dear Mukal,

A,B,C,E (red dots) will represents the centres of sphere of atoms in HCP closed packing,

now, AB=BC=CA=EB = 2R (edge lengths of HCP)

BD=\frac{2R}{\sqrt{3}}  ,as BD divides the median through B in 2:1 ratio from B...

now EDB is rt. triangle,

by pythagores theorem :

(\frac{2R}{\sqrt{3}})^2+h^2=(2R)^2

=> h=2R\sqrt{\frac{2}{3}}

now, their will be another atom which will be mirror image of E on plane ABC

so, the height (c) of HCP will be 2h

c=2h=4R\sqrt{\frac{2}{3}}

Best Of luck

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert


Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...