The correct relation between interatomic force constant $K$ . Young modulus $Y$ and interatomic distance ${r_0}$ is:
A) $K = Y{r_0}$
B) $K = \dfrac{{{r_0}}}{Y}$
C) $K = \dfrac{Y}{{{r_0}}}$
D) $K = {r_0}^3{Y^2}$

Hint: The interatomic force constant is the ratio of the interatomic force and the change in interatomic distance. Assuming a solid object made up of many atoms behaves like springs, we may conclude the concept of expansion of spring due to interatomic force. Find the effective force constant for all the springs over there in that object which will be the product of the force constant for one pair of atoms in the spring and the ratio of the number of springs in parallel and series position. From the above relation, you will find the effective force constant (interatomic force constant) in terms of force, expansion of length. Use the formula for young’s modulus and find the relation between the interatomic force constants, Young modulus, and interatomic distance

Formula used:
The interatomic force-constant $K$ = $\dfrac{F}{{\Delta{r_0}}}$
$F$ = The interatomic force,
$\Delta{r_0}$ = the change in interatomic distance.
$L$ = the length of the object and $\Delta L$ is the elongation due to the force.
Hence,$K = \dfrac{F}{{\Delta L}}$
The effective Force constant for all the spring, $K = \dfrac{{{N_{parallel}}}}{{{N_{series}}}}K'$ where, $K'$ is the interatomic force constant of one spring.
${N_{parallel}}$ = the number of spring in the parallel position = $\dfrac{A}{{{r_0}^2}}$
$A$ = area of the surface
${r_0}$ = the interatomic distance
${N_{series}}$ = the number of spring in the series position = $\dfrac{L}{{{r_0}}}$
$\dfrac{F}{{\Delta L}} = \dfrac{{{N_{parallel}}}}{{{N_{series}}}}K'$
The young’s modulus, $Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$.

Complete step by step answer:
A solid object is of length $L$ is expanded due to a force. The interatomic force- constant $K$ = $\dfrac{F}{{\Delta {r_0}}}$
$F$ = The interatomic force,
$\Delta {r_0}$ = the change in interatomic distance.
If we assume that the object is made up of atoms those are behaving like springs, and the amount of expansion is $\Delta L$,
The interatomic force constant $K$ is written as, $K = \dfrac{F}{{\Delta L}}$
Since there are many springs inside the object.
The effective Force constant for all the spring,$K = \dfrac{{{N_{parallel}}}}{{{N_{series}}}}K'$ where, $K'$ is the force constant of one spring.
The area,$A = {N_{parallel}} \times {r_0}^2$ , ${r_0}$ = the interatomic distance
${N_{parallel}}$ = the number of spring in the parallel position
$\therefore {N_{parallel}} = \dfrac{A}{{{r_0}^2}}$
The length, $L = {N_{series}} \times {r_0}$
${N_{series}}$ = the number of spring in the series position
$\therefore {N_{series}} = \dfrac{L}{{{r_0}}}$
So, $K = \dfrac{{{N_{parallel}}}}{{{N_{series}}}}K'$
$ \Rightarrow K = \dfrac{{\dfrac{A}{{r_0^2}}}}{{\dfrac{L}{{{r_0}}}}}K'$
$ \Rightarrow K = \dfrac{A}{{L{r_0}}}K'$
Since, $K = \dfrac{F}{{\Delta L}}$
$\therefore \dfrac{F}{{\Delta L}} = \dfrac{A}{{L{r_0}}}K'$
$ \Rightarrow K' = \dfrac{{FL{r_0}}}{{A\Delta L}}$
We know, The young’s modulus, $Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$ = $\dfrac{{FL}}{{A\Delta L}}$
Hence, $K' = Y \times {r_0}$
So The interatomic force constant can be written from the above concept, $K = Y{r_0}$ .

Hence the right answer is in option \[\left( A \right)\].

Note: We assume that the object is made up of many springs. The Area of the object is represented by the product of the total number of springs in parallel position and the area of a unit square. And the length of the object is defined as the product of the number of springs in series position and the distance between them. Since the effective force constant of spring is directly proportional to the number number of spring in parallel position and inversely proportional to the number of spring in series position, we take the ratio of ${N_{parallel}}$ as numerator and ${N_{series}}$ as the denominator.