How much kinetic energy must a deuterium ion (charge = 1.6 x 10^-19 C)have to approach within 1.0 x 10^-14m of another deuterium ion? (k_e =8.99 x 10⁹ N*m²/C² and 1 keV = 1.6 x 10^-16 J)

Question

Quizplus · Accepted Answer

The electric potential energy between two deuterium ions with charge $q$ separated by distance $r$ is given by:
$$U_e=\frac{k_e q^2}{r}$$
Since the two ions will repel each other, the total energy of the system (kinetic + potential) must be positive in order for the ions to approach each other. Therefore, the kinetic energy $K$ required is the difference between the total energy and the potential energy at separation distance $r$:
$$K = U_{tot} - U_e = U_{tot} - \frac{k_e q^2}{r}$$
where $U_{tot}$ is the total energy of the system.
The problem doesn't provide the total energy, but it does give a distance of $r=10^{-14}$ m. Let's assume that at this distance the potential energy is equal to the kinetic energy required to approach even closer (i.e. we neglect the difference between the potential energy at $r$ and infinity). Then:
$$K = \frac{k_e q^2}{r} = \frac{8.99	imes10^9\,\mathrm{N\cdot m^2/C^2}	imes(1.6	imes10^{-19}\,\mathrm{C})^2}{10^{-14}\,\mathrm{m}} \approx 144\,\mathrm{keV}$$
Therefore, the best choice is D, 140 keV.

How Much Kinetic Energy Must a Deuterium Ion (Charge =