91Ó°ÊÓ

The force of attraction between two charged particles kept in the vicinity is called the Coulombic force of attraction. The magnitude of the force depends on the product of the two charges and inversly proporional to the distance between them. The Coulombic force of attraction is given by

$\mathit{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, r is the distance between the two charged particles, and q₁and q₂ are the charges.

It’s been given that the alpha particle contains two protons and two neutrons; hence it has a net charge of +2e.

It is also given that the single proton has a net charge of -e.

The distance between the alpha particle and the proton is 1 mm.

From Coulomb's law, we know that

$F=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r^2}$

Substitute the values and solve as below.

$$\begin{gathered} F=9×{10}^9×\frac{2e×e}{r^2} \\ =9×{10}^9×\frac{2{\left(1.6×{10}^{-19}\right)}^2}{{\left(1×{10}^{-3}\right)}^2} \\ =4.608×{10}^{-22}\text{N} \end{gathered}$$

It’s been given that the alpha particle contains two protons and two neutrons; hence it has a net charge of +2e.

It is also given that the single proton has a net charge of -e.

The distance between the alpha particle and the proton is 1 mm.

From Coulomb's law, we know that

$F=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r^2}$

Substitute the values and solve as below.

$$\begin{gathered} F=9×{10}^9×\frac{2e×e}{r^2} \\ =9×{10}^9×\frac{2{\left(1.6×{10}^{-19}\right)}^2}{{\left(1×{10}^{-3}\right)}^2} \\ =4.608×{10}^{-22}\text{N} \end{gathered}$$

Therefore, the forces are equal in magnitude. The correct option is (c): The forces are equal in magnitude.