Question

It is possible to estimate the activation energy for fusion by calculating the energy required to bring two deuterons close enough to one another to form an alpha particle. This energy can be obtained by using Coulomb's law in the form \(E=8.99 \times 10^{9} \mathrm{q}_{1} q_{2} / r,\) where \(q_{1}\) and \(q_{2}\) are the charges of the deuterons \(\left(1.60 \times 10^{-19} \mathrm{C}\right), r\) is the radius of the \(\mathrm{He}\) nucleus, about \(2 \times 10^{-15} \mathrm{~m},\) and \(E\) is the energy in joules. (a) Estimate \(E\) in joules per alpha particle. (b) Using the equation \(E=m v^{2} / 2\), estimate the velocity (meters per second) each deuteron must have if a collision between the two of them is to supply the activation energy for fusion \((m\) is the mass of the deuteron in kilograms).

Short answer

Answer: The required velocity for each deuteron is approximately \(2.23 \times 10^6 \, \text{m/s}\) for the collision to provide enough activation energy for fusion.

Step-by-step solution

Calculate the energy (E) using Coulomb's law

First, we use Coulomb's law to find the energy required to bring two deuterons close enough to form an alpha particle. The formula is given as \(E=\frac{8.99 \times 10^9 \times q_1 q_2}{r}\). We know that \(q_1=q_2=1.60 \times 10^{-19} \, \text{C}\) and \(r=2 \times 10^{-15} \, \text{m}\). Plug in the values and solve for E: \(E=\frac{8.99 \times 10^9 \times (1.60 \times 10^{-19})^2}{2 \times 10^{-15}}\)

Estimate \(E\) in Joules per alpha particle

Now that we have the formula set, we can solve for the energy as following: \(E=\frac{8.99 \times 10^9 \times (1.60 \times 10^{-19})^2}{2 \times 10^{-15}}= 1.15 \times 10^{-13} \, \text{J}\) The energy, E, per alpha particle is approximately \(1.15 \times 10^{-13} \, \text{J}\).

Estimate the velocity (v) each deuteron must have

Now, we have to estimate the velocity (v) needed by the deuterons if the collision supplies the activation energy for fusion. To do this, we will use the energy formula \(E=\frac{m v^2}{2}\). Let's first solve for v: \(v=\sqrt{\frac{2E}{m}}\) We know the energy (E) per alpha particle is about \(1.15 \times 10^{-13} \, \text{J}\), and the mass of a deuteron (m) is about \(3.34 \times 10^{-27} \, \text{kg}\). Plug in the values and solve for v: \(v=\sqrt{\frac{2(1.15 \times 10^{-13})}{3.34 \times 10^{-27}}}= 2.23 \times 10^6 \, \text{m/s}\) Each deuteron must have a velocity of approximately \( 2.23 \times 10^6 \, \text{m/s}\) for the collision to provide enough activation energy for fusion.