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I am in my first year of college and doing an EPQ type project on nuclear fusion and am wondering about Coulomb's Law.

I am looking into the force experienced by a particle due to the Coulomb force and would like to find an equation that describes how much energy is needed to overcome this Coulomb force so the two hydrogen atoms can fuse.

I have come up with an equation that makes sense to me and describes the amount of energy required for the two atoms to fuse. However, I'm uncertain if another equation I've found describes it better.


This is my thinking:

If Coulomb's law is given by the equation

$$F = \frac{k \ q_1 q_2 \ e^2}{r^2}$$

then isn't the energy needed to overcome the repulsion force the integral of that equation between two distances $r_1$ and $r_2$ where where $r_2$ would preferably be smaller than the barrier of the strong nuclear force and $r_1$ is any distance that the particle starts from:

$$E = \int_{r_2}^{r_1} \frac{q_1 q_2 \ e^2}{ 4 \pi \xi_0 r^2} dr$$

If it is then what does this equation describe? It is the coulomb potential energy equation. $$ V_C=\frac{e^2}{4\pi\epsilon_0}\frac{Z_aZ_b}{R_a+R_b}$$

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Your equation for $E$, the work done by an external force to bring the two charges from a separation of $r_2$ to a separation of $r_1$ should be

$$E = \int_{r_2} ^{r_1} -\dfrac {q_1q_2 e^2}{4 \pi \epsilon_0 r^2 } dr = \dfrac {q_1q_2 e^2}{4 \pi \epsilon_0 } \left( \dfrac {1}{r_1}-\dfrac {1}{r_2}\right)$$

If you take the electric potential energy $E$ to be zero when the separation of the charges at the beginning $r_2 =\infty$ then this equation reduces to

$$E = \dfrac {q_1q_2 e^2}{4 \pi \epsilon_0 r_1} $$

and this is to be compared with your equation $$V_{\rm c} E = \dfrac {Z_{\rm a} Z_{\rm b} e^2}{4 \pi \epsilon_0 (R_{\rm a} +R_{\rm b})} $$ with $q_1 = Z_{\rm a}, \, q_2 = Z_{\rm b}$ and $r_1 = R_{\rm a}+R_{\rm b}$ the closest approach of the centres of the two nucleii where $R_{\rm a}$ and $R_{\rm b}$ are the radii of the hydrogen nucleii.

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    $\begingroup$ in addition, you can estimate the nuclei-radius by $R = r_0A^{1/3}$, where $r_0\approx 1.3\cdot10^{-15}$ and $A$ the mass number, and insert. $\endgroup$ – Alf Mar 8 '18 at 17:56

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