Question

Suppose the magnitude of Nuclear force between two protons varies with the distance between them as shown in figure. Estimate the ratio "Nuclear force/Coulomb force" for
(a) x = 8 fm
(b) x = 4 fm
(c) x = 2 fm
(d) x = 1 fm (1 fm = 10 ⁻¹⁵m).
Figure

Answer 1

First let us calculate the coulomb force between 2 protons for distance = 8 fm
$$\begin{gathered} F\mathit{}\mathit{=}\frac{\mathit{K}\mathit{}{\mathit{q}}^{\mathit{2}}}{{\mathit{r}}^{\mathit{2}}} \\ =\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{(8\times {10}^{-15}{)}^2} \\ =3.6\mathrm{N} \end{gathered}$$
$$\begin{gathered} F_{\mathrm{N}}=0.05\mathrm{N} \\ \frac{F_{\mathrm{N}}}{F_{\mathrm{C}}}=\frac{0.05}{3.6}=0.0138\mathrm{N} \end{gathered}$$
For x= 4 fm
$$\begin{gathered} F_{\mathrm{C}}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{(4\times {10}^{-15}{)}^2} \\ =\frac{23.04\times {10}^{-29}}{(4\times {10}^{-15}{)}^2} \\ =14.4\mathrm{N} \\ {F}_{\mathrm{N}}=1\mathrm{N} \\ \frac{F_{\mathrm{N}}}{F_{\mathrm{C}}}=\frac{1}{14.4}=0.0694\mathrm{N} \\ \mathrm{For}x=2\mathrm{fm} \\ {F}_{\mathrm{C}}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{(2\times {10}^{-15}{)}^2} \\ =57.6\mathrm{N} \\ {F}_{\mathrm{N}}=10\mathrm{N} \\ \frac{F_{\mathrm{N}}}{F_{\mathrm{C}}}=\frac{10}{57.6}=0.173 \\ \mathrm{For}x=1\mathrm{fm} \\ {F}_{\mathrm{C}}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{(1\times {10}^{-15}{)}^2} \\ =230.4\mathrm{N} \\ {F}_{\mathrm{N}}=1000\mathrm{N} \\ \frac{F_{\mathrm{N}}}{F_{\mathrm{C}}}=\frac{1000}{230.4}=4.34 \end{gathered}$$