Question 41.7: CASE STUDY The Probability of Proton Tunneling in the Sun Eq......
CASE STUDY The Probability of Proton Tunneling in the Sun
Equation 41.22
P_{\text {tunnel }} \approx e^{-\gamma} \quad \quad (41.22)is for a rectangular barrier. Although the barrier in Figure 41.18 is not rectangular, use Equation 41.22 to make an order-of-magnitude estimate of the probability of tunneling required for two protons to fuse together in the Sun. (Don’t be surprised if your estimate is off.) Note that in Figure 41.18, x_0 \sim 10^{-15} m (approximate diameter of a proton).
INTERPRET and ANTICIPATE
We expect to find a small but non-zero result.
SOLVE
We need the barrier potential energy and the initial kinetic energy in SI units.
It is also helpful to find their difference.
\begin{aligned}& U_b-K_i \sim\left(10^{-13} J -10^{-16} J \right) \\& U_b-K_i \sim 10^{-13} J\end{aligned}The strong force is important at x_0, so take this as the left edge of the barrier.
x_0 \sim 10^{-15} mAt the right edge of the barrier, the proton’s initial kinetic energy equals the system’s potential energy given by Coulomb’s law.
\begin{aligned}& \frac{k e^2}{x}=K_i \\& x=\frac{k e^2}{K_i}=\frac{\left(9 \times 10^9 N \cdot m ^2 / C ^2\right)\left(1.6 \times 10^9 C \right)^2}{10^{-16} N \cdot m } \\& x \sim 10^{-12} m\end{aligned}If we simply subtract the left edge x_0 from the right edge x, we get an overestimate for the width of the barrier because the barrier is not a rectangle.
x-x_0=\left(10^{-12} m -10^{-15} m \right)=10^{-12} mA slightly better estimate for the average width of the barrier is about an order of magnitude smaller.
w \sim 10^{-13} mNow we can use Equation 41.23 to estimate γ. (The subscript p on the m stands for proton.)
\begin{aligned}& \gamma^2 \equiv \frac{32 \pi^2 m_p w^2\left(U_b-K_i\right)}{h^2} \quad \quad (41.23)\\ & \gamma^2 \sim \frac{(300)\left(10^{-27} kg \right)\left(10^{-13} m \right)^2\left(10^{-13} J \right)}{\left(7 \times 10^{-34} J \cdot s \right)^2} \sim 600 \\& \gamma \sim 25\end{aligned}The probability of tunneling comes from Equation 41.22
\begin{aligned}& P_{\text {tunnel }} \approx e^{-\gamma} \\& P_{\text {tunnel }} \sim e^{-25} \sim 10^{-11}\end{aligned}CHECK and THINK
We modeled the barrier as rectangular. A better model finds a higher probability of about P_{\text {tunnel }} \sim 10^{-10}. It means that if 10^{10} pairs of protons attempt to tunnel, only one pair is successful. This might sound like a very low probability. However, it is a high enough probability to account for the 10^{38} fusion reactions (see Example 39.10, page 1295) that take place every second in the core of the Sun.