Answer
$
r_{\min }=1pm$
Work Step by Step
The kinetic energy of each proton is
$
K=k_B T=\left(1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)\left(1.0 \times 10^7 \mathrm{~K}\right)=1.38 \times 10^{-16} \mathrm{~J} .
$
At the closest separation, $r_{\min }$, all the kinetic energy is converted to potential energy:
$
K_{\text {tot }}=2 K=U=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_{\min }} .
$
Solving for $r_{\min }$, we obtain
$
r_{\min }=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 K}=\frac{\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right)\left(1.60 \times 10^{-19} \mathrm{C}\right)^2}{2\left(1.38 \times 10^{-16} \mathrm{~J}\right)}=8.33 \times 10^{-13} \mathrm{~m} \approx 1 \mathrm{pm} .
$
