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A given metal crystallises in BCC structure of radius of metal atom is 120 pm, then the distance between two next nearest neighbour is -

  • Option 1)

    200 pm

  • Option 2)

    277 pm

  • Option 3)

    382 pm

  • Option 4)

    240 pm

 

Answers (1)

In BCC structure distance between two nearest neihghbour is \frac{a\sqrt{3}}{2} and distance between two next neighbour is a.

radius of atom r=\frac{a\sqrt{3}}{4}

a=\frac{4r}{\sqrt{3}}=\frac{4\times 120}{1.732}=277pm

 

Relation between radius of constituent particle, R and edge length, a for body centered cubic unit cell -

-

 

 


Option 1)

200 pm

This is incorrect

Option 2)

277 pm

This is correct

Option 3)

382 pm

This is incorrect

Option 4)

240 pm

This is incorrect

Posted by

Vakul

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