If the inter-atomic spacing for a metal is 2.8 Å and its Young's modulus is `2 xx 10^(11) Nm^(-2)` then for a force of `10^(9) Nm^(-2)` the inter-atomic force constant is

Options

A. `56 Nm^(-1)`
B. `5.6 Nm^(-1)`
C. `Nm^(-1)`
D. `6.5 Nm^(-1)`

Answer

A

Explanation

Let `r_(0)` be the interatomic distance then area of a chain of chain of atoms can be taken as `r_(0)xx r_(0)= r_(0)^(2)`
Using `Y= (F)/(A) (l)/(Deltal) ` we get
`Y = (F)/(r_(0)^(2)) (r_(0))/(Deltal)= (F)/(Deltal) (r_(0))/(r_(0)^(2))= (k)/(r_(0)) ` or ` k = Yr_(0) = 2xx 10^(11) xx 2.8 xx 10^(-10) = 56 Nm^(-1)`