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Nuclear Fusion in the Sun. The source of the sun’s energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact. (a) Assume both protons are moving with the same speed and they collide head on. If the radius of the proton is 1.2 * 10^-15m, what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is 1.67 * 10^-27kg. (b) Another nuclear fusion reaction that occurs in the sun’s core involves a collision between two helium nuclei, each of which has 2.99 times the mass of the proton, charge +2e, and radius 1.7 * 10^-15m. Assuming the same collision geometry as in part (a), what minimum speed is required for this fusion reaction to take place if the nuclei must approach a center to­ center distance of about 3.5 * 10^-15m? As for the proton, the charge of the helium nucleus is uniformly distributed throughout its volume. (c) In Section 18.3 it was shown that the average translational kinetic energy of a particle with mass m in a gas at absolute temperature T is 3/2 kT, where k is the Boltzmann constant (given in Appendix F). For two protons with kinetic energy equal to this average value to be able to undergo the process described in part (a), what absolute temperature is required? What absolute temperature is required for two average helium nuclei to be able to undergo the process described in part (b)? (At these temperatures, atoms are completely ionized, so nuclei and electrons move separately.) (d) The temperature in the sun’s core is about 1.5 * 10⁷ K. How does this compare to the temperatures calculated in part (c)? How can the reactions described in parts (a) and (b) occur at all in the interior of the sun? (Hint: See the discussion of the distribution of molecular speeds in Section 18.5.)

Step by step solution

01

Step 1:

According to conservation law of energy:

$\mathbf{∆}\mathbf{E}\mathbf{-}\mathbf{∆}\mathbf{U}\mathbf{=}\mathbf{0}$

role="math" localid="1664267011369" $$\begin{gathered} ∆\mathrm{E}-∆\mathrm{U} \\ 2\mathrm{E}=∆\mathrm{U} \\ 2\mathrm{E}=\frac{{\mathrm{Ke}}^2}{\mathrm{r}} \\ 2×\frac{1}{2}{\mathrm{mv}}^2=\frac{{\mathrm{Ke}}^2}{\mathrm{r}} \\ \mathbf{v}\mathbf{=}\sqrt{\frac{{\mathbf{Ke}}^{\mathbf{2}}}{\mathbf{2}\mathbf{mr}}} \end{gathered}$$

02

Step 2:

The radius of the proton is r_p=1.2 x 10^-15 m and the mass of the proton m_p=1.67x 10^-27 kg.

Solve for the minimum speed that will allow fusion to occur:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{{\mathrm{Ke}}^2}{2{\mathrm{m}}_{\mathrm{p}}\mathrm{r}}} \\ =\sqrt{\frac{8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2{\left(1.62×{10}^{-19}\mathrm{C}\right)}^2}{2×1.67×{10}^{-27}\mathrm{kg}×1.2×{10}^{-15}\mathrm{m}}} \\ =7.672×{10}^6\mathrm{m} \end{gathered}$$

So, the minimum speed that will allow fusion to occur is 7.672 x 10⁶ m/s.

03

Step 3:

The radius of the helium is r_He=3.5 x 10^-15 m,mass of the heium m_He=2.99 x 10^-27 kg and number of electrons=4

Solve for the minimum speed that will allow fusion to occur:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{{\mathrm{Ke}}^2}{2{\mathrm{m}}_{\mathrm{He}}{\mathrm{r}}_{\mathrm{He}}}} \\ =\sqrt{\frac{8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2{\left(4×1.62×{10}^{-19}\mathrm{C}\right)}^2}{2×2.99×1.67×{10}^{-27}\mathrm{kg}×3.5×1.2×{10}^{-15}\mathrm{m}}} \\ =1.039×{10}^7\mathrm{m} \end{gathered}$$

So, the minimum speed that will allow fusion to occur is 1.039 x 10⁷ m/s.

04

Step 4:

According to conservation law of energy:

$$\begin{gathered} ∆\mathrm{E}-∆\mathrm{U} \\ \frac{1}{2}{\mathrm{mv}}^2=\frac{3}{2}\mathrm{kT} \\ \mathbf{T}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{3}\mathbf{k}} \end{gathered}$$

Solve for the absolute temperature that will allow fusion to occur (for (a)):

$$\begin{gathered} \mathrm{T}=\frac{{\mathrm{m}}_{\mathrm{p}}{\mathrm{v}}^2}{3\mathrm{k}} \\ =\frac{1.67×{10}^{-27}\mathrm{kg}{\left(7.672×{10}^7\mathrm{m}/\mathrm{s}\right)}^2}{3×1.3807×{10}^{-23}\mathrm{J}/\mathrm{K}} \\ =2.373×{10}^9\mathrm{K} \end{gathered}$$

The absolute temperature that will allow fusion to occur is$2.373×{10}^9\mathrm{K}$

05

Step 5:

Solve for the absolute temperature that will allow fusion to occur (for (b)):

$$\begin{gathered} \mathrm{T}=\frac{{\mathrm{m}}_{\mathrm{He}}{\mathrm{v}}^2}{3\mathrm{k}} \\ =\frac{2.99×1.67×{10}^{-27}\mathrm{kg}{\left(1.039×{10}^7\mathrm{m}/\mathrm{s}\right)}^2}{3×1.3807×{10}^{-23}\mathrm{J}/\mathrm{K}} \\ =1.3014×{10}^{10}\mathrm{K} \end{gathered}$$

The absolute temperature that will allow fusion to occur is$1.3014×{10}^{10}\mathrm{K}$

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