How fast must two deuterium atoms be moving so they can overcome the Coulomb force of repulsion, and attain the necessary 10−¹⁴ m for fusion? (m( ) = 2.014 1 u)

Question

Quizplus · Accepted Answer

The kinetic energy required is equal to the electrostatic (Coulombic) potential energy when the two deuterium nuclei are separated by a distance of $10^{-15}$ meters:
$$\frac{1}{2}mv^2 = \frac{kQ_1Q_2}{r}$$
where $m$ is the mass of a deuterium nucleus, $v$ is the velocity of a deuterium nucleus, $k$ is Coulomb's constant, $Q_1$ and $Q_2$ are the charges of the two deuterium nuclei, and $r$ is the separation distance between the two nuclei.

For deuterium, $Q_1=Q_2=e$ (the charge of a proton) and $m=2.0141$ u. Plugging in these values and solving for $v$ gives:
$$v = \sqrt{\frac{2kQ_1Q_2}{m r}} = \sqrt{\frac{2k e^2}{m r}}$$
Substituting $r=10^{-15}$ m, we get:
$$v = \sqrt{\frac{2 \cdot 8.99	imes10^9 \cdot (1.60	imes10^{-19})^2}{(2.0141	imes 1.66	imes10^{-27}) (10^{-15})}} \approx 2.6	imes 10^6	ext{ m/s}$$
Therefore, the answer is (C) $2.6 	imes 10^{6}$ m/s.

How Fast Must Two Deuterium Atoms Be Moving So They