Question

A helium ion (He⁺⁺) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 * 10^-15 m, and e is the magnitude of the charge of an electron or a proton.) (a) 2e; (b) 11e; (c) 20e; (d) 22e.

Short answer

11e charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus.

Step-by-step solution

Potential energy

Potential energy between charges is given by the equation

${\mathbf{U}}_{\mathbf{b}}\mathbf{=}\mathbf{k}\frac{{\mathbf{qq}}_{\mathbf{He}}}{\mathbf{r}}$

Determine the charge of the nucleus

The potential energy between two particles is given by

${\mathbf{U}}_{\mathbf{b}}\mathbf{=}\mathbf{k}\frac{{\mathbf{qq}}_{\mathbf{He}}}{\mathbf{r}}$

Helium has the charge of two protons

So, the next equation will be

$$\begin{gathered} {\mathbf{K}}_{\mathbf{a}}\mathbf{+}{\mathbf{U}}_{\mathbf{a}}\mathbf{=}{\mathbf{K}}_{\mathbf{b}}\mathbf{+}{\mathbf{U}}_{\mathbf{b}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}{\mathbf{U}}_{\mathbf{b}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{k}\frac{{\mathbf{qq}}_{\mathbf{He}}}{\mathbf{r}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{q}\mathbf{=}\frac{{\mathbf{rk}}_{\mathbf{a}}}{{\mathbf{kq}}_{\mathbf{He}}} \end{gathered}$$

Now from the equation the charge is

$$\begin{gathered} \mathbf{q}\mathbf{=}\frac{{\mathbf{rK}}_{\mathbf{a}}}{\mathbf{2}\mathbf{ek}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\left(10\times {10}^{-15}m\right)\left(3.0\times {10}^6\mathrm{ev}\times 1.6\times {10}^{-19}J/\mathrm{eV}\right)}{\mathbf{2}\left(1.6\times {10}^{-19}J\right)\left(9.0\times {10}^{9}N.m^2/C^2\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{66}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{C} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{11}\mathbf{e} \end{gathered}$$

Therefore, 11e charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus.