Solid State

its nowhere written that the central large sphere is hollow and y did u take R=10r

ya,
then it may be
because ,otherwise the question was not clear
well.

Ya Pratik this is actually a qs from a comprehension

I had a doubt to use the general P.f. & so i posted

the diagram has given a completely new angle to da problem... as hcp is not working guess yu should start from da basics... can yu pl paste da exact wordings of da question as HOLLOW sphere is something i cant relate to. even then if yu say tht there is an imaginary (hollow) sphere and small spheres around,how are they packed??(does da question give hcp?if it does, i see a contradiction to da dimensions given) pl put the passage here...

If the diagram is as given by Ashish ,
then there may be two cases :
1) large sphere is hollow
2) large sphere is solid

if large sphere is hollow then answer must be 0.002.
if not then there is a problem because ,

p.f =

will be greater than 1.
which can not be.
it may be 1 but for that case ,

otherwise question is not clear.

post the full paragraph here. if otherwise.

well.. this was a ques from aakash aits.. so I and Uttara know the complete ques (as it was a paragraph type) which mentions that the large sphere is hollow and its radius R = 10 times the radius of the smaller spheres..

even then (i slightly doubt) there is something wrong with my method.
Here bcc formula that a√3=2(r+R) CANNOT be applied. (as then the large sphere (for R=10r) will not remain completely inside the cube

why is R=10r??

my soln:
if u take spheres at corners to be negligible and that da central sphere is very large fitting completely in da cube then:
a=2R
PF=vol of central sphere / vol of cube
=(4/3)pi R³a³
hence, pf=0.52
option(C)

the two spheres are of different radii and so R=10r cannot be taken.
if we conder it as a spherewhich is solid
the two radii being r & R then by approaching as in bcc n taking
a√3=2(r+R) n applying formulae the p.f. comes as

√32*pi

which is equal to somthing like 2.7

so either the sum is wrong or else it has to looked at with a different view

but the other structures doesnot satisfy the given data

so uttora should re chek her sum

(D) 0.002

Packing fraction depends on the volume (solid volume).. So contribution of hollow sphere to the PF is negligible

So total volume of smaller spheres in the cube = 8*1/8 (corner atoms so contribution of each to one unit cell = 1/8)*V_sph = V_sph

Now as it can be treated as BCC,

a√3 = 2(R+r)

So u can get volume of cube as a³

PF = total volume of smaller spheres in the cube/volume of cube
putting the values and taking R=10r

we get PF = 0.002

So how can the height be found from this structure of hcp??

When 0.68 packing fraction was calculated,,all spheres were of same size,,here its not,,so ans is bound to be different,,and to find it we have to go from the basics like asish did.

in bcc we have all the spheres were of same size that is they have the same size.But here it is different.Wat asish did is perfectly rite,and i cant understand wat he means by hollow.

In a hcp structure, are the spheres always linked to each other??

it is no where said that is not a solid sphere...

ashish this is a case of bcc

@Uttara: BCC waala answer is valid ONLY when the body centre sphere is completely solid.. Here the body centre sphere is NOT completely solid (it is hollow) so you have to work from the basics only

@Asish This method I have seen in soln But for a BCC structure P.F. is 0.68 na??

& this is clearly a BCC