MCQ 11 Mark
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$x = 8fm$
- B$x = 4fm$
- C$x = 2fm$
- D$x = 1fm (1fm = 10^{-15} m).$
Answer
View full question & answer→First let us calculate the coulomb force between 2 protons for distance:
$ =3.6 \mathrm{NFN}$
$ =0.05 \mathrm{NFNFC}=0.053 .6=0.0138 \mathrm{~N}$
$= 23.04 × 10 - 29(4 × 10 - 15) = 14.4NF$
$= 1NFNFC = 114.4 = 0.0694N$
$= 57.6 NFN = 10 NFNFC = 1057.6 =0.173$
$= 230.4 NFN = 1000 NFNFC = 1000230.4 = 4.34$
- $x = 8fm$
$ =3.6 \mathrm{NFN}$
$ =0.05 \mathrm{NFNFC}=0.053 .6=0.0138 \mathrm{~N}$
- $x = 4fm$
$= 23.04 × 10 - 29(4 × 10 - 15) = 14.4NF$
$= 1NFNFC = 114.4 = 0.0694N$
- $x = 2fm$
$= 57.6 NFN = 10 NFNFC = 1057.6 =0.173$
- $x = 1fm$
$= 230.4 NFN = 1000 NFNFC = 1000230.4 = 4.34$