Silver metal crystallises in a cubic closed packed arrangement with edge length 404 pm .Thus radius of the silver atom is

Options

A. `203.5` pm
B. `138.8` pm
C. `142.8` pm
D. `407.0` pm

Answer

C

Explanation

`AB = a, AC = a`

`BC^(2) = AB^(2) + AC^(2) = 2a^(2)`
Radius `= r_(1)`
then `BC - CD + DE + EB`
`= r + 2r+ r= 4r`
`16r^(2) = 2a^(2)`
`r^(2) = (a^(2))/(B)`
`r = (a)/(2sqrt(2)) = (404)/(2 sqrt(2)) = (404)/(2.828) = 142.8` pm