Question

Suppose an attractive nuclear force acts between two protons, which may be written as $F=\frac{{Ae}^{-kr}}{r^2}$. The dimensional formula for $\frac{A}{K}$ is

• A. $M^1{L}^2{T}^2$
• B. $M^1{L}^3{T}^{-1}$
• C. $M^1{L}^4{T}^1$
• D. $M^1{L}^4{T}^{-2}$

Answer 1

Answer: (D)

$M^1{L}^4{T}^{-2}$

Given,

$F=\frac{A{e}^{-kr}}{r^2}$

$\left[F\right]=\frac{\left[A\right]}{\left[r^2\right]}$

$⟹\left[A\right]=\left[F\right]\left[r^2\right]$

$⟹\left[A\right]=\left[ML{T}^{-2}\right]\left[L^2\right]$

$\left[A\right]=\left[M{L}^3{T}^{-2}\right]$

We know that $Kr$ is unitless.

Therefore, S.I unit of $K=\frac{1}{m}$

Then,

$\left[K\right]=\frac{1}{\left[L\right]}=[L^{-1]}$

So,

$\frac{\left[A\right]}{\left[k\right]}=\frac{\left[M{L}^3{T}^{-2}\right]}{\left[L^{-1}\right]}=\left[M^1{L}^4{T}^{-2}\right]$