Home
Class 12
CHEMISTRY
Silver crystallises in fcc latice. If ed...

Silver crystallises in fcc latice. If edge length of the unit cell is `4.077xx10^(-8)cm`, then calculate the radius of silver atom.

Text Solution

Verified by Experts

For FCC `a=2sqrt(2)r`
`r=(4.077xx10^(-8))/(2xx1.414)=1.44xx10^(-8)cm`
Promotional Banner

Topper's Solved these Questions

  • XII BOARDS

    XII BOARDS PREVIOUS YEAR|Exercise SET-II|32 Videos

  • XII BOARDS

    XII BOARDS PREVIOUS YEAR|Exercise SET-III|34 Videos

  • SAMPLE PAPER 2019

    XII BOARDS PREVIOUS YEAR|Exercise SECTION: D|1 Videos

Similar Questions

Explore conceptually related problems

Silver crystallizes in fcc lattic. If the edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 g cm^(-3) . Calculate the atomic mass of silver.

Sodium crystallizes in bcc structure. If the edge length of unit cell is 4.3xx10^-s cm, the radius of Na atom is

Silver crystallises in a fcc lattice. The edge length of its unit is 4.077xx10^(-8)cm and its density is 10.5g cm^(-3) . Claculate on this basis of the atomic mass of silver (N_(A)=6.02xx10^(23) "mol"^(-1))

Silver metal crysatllises with a face centred cubic lattice. The length of the unit cell is found to be 4.077xx10^(-8)cm . Calulate the atomic radius and density of silver.

Polonium crystallises in a simple cubic structure. If the edge length of the unit cel is 336 pm, calculate the atomic radius of polonium.