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Born–Oppenheimer approximation - Wikipedia

In quantum chemistry and molecular **physics**, the Born–Oppenheimer (BO)
approximation is the best known mathematical approximation in molecular
dynamics. Specifically, it is the assumption that the wave functions of atomic
**nuclei** and ... The slope of the **potential energy** surface can be **used** to simulate
molecular ...

For more information, see Born–Oppenheimer approximation - Wikipedia

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I found an answer from ntrs.nasa.gov

Proceedings of the **Second** International Colloquium on

Mar 1, 1982 **...** their **common** problenls and learn from **each other**. hi the intervening seven ...
this shape **can** be **calculated** from knowledge of the surface energy as a ... The
interaction effects **between two** bubbles migrating along their line of centers
under ... In physical terms, in part, **it** is **assumed** that the vorticity and.

For more information, see Proceedings of the **Second** International Colloquium on

**Coulomb's law **states that the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

F = \frac{1}{4 \pi \epsilon_0} * \frac{q_1q_2 }{d^2} , where \text{ constant } \frac{1}{4 \pi \epsilon_0} = 9 * 10^{9} Nm^2/C^2 , q_1 \text{ and } q_2 - charges and d - distance between the two chargs

Given that

Radius of the deuteron r = 2.0 femtometer (fm)

r = 2.0 * 1* 10^{-15} = 2 * 10^{- 15} meter. 1 fm = 1* 10^{-15}

Charge of a deuterons (proton) q_1 = q_2 = 1.6 * 10^{-19} C

Step 1: Calculating the potential energy of two charged bodies

distance between the two deuterons d = Radius of the deuteron + Radius of the deuteron

d = 2 * 10^{15} + 2 * 10^{15} = 4 * 10^{-15} m

Potential energy P.E = \frac{1}{4 \pi \epsilon_0} * \frac{e^2}{d}

P.E = \frac{1}{4 \pi \epsilon_0} * \frac{(1.6 * 10^{-19})^2}{(4 * 10^{-15})} Joules

P.E = \frac{9 * 10^{9}}{1.6 * 10^{-19}} * \frac{(1.6 * 10^{-19})^2}{(4 * 10^{-15})} eV 1 eV = 1.6 * 10^{-19} J

P.E = 360 KeV

**Hence, height of the two-deuteron system's potential barrier ** P.E = 360 KeV

Step 2: Determining the kinetic energy of two charged bodies

Recall the relation between the kinetic(K.E) and potential energy (P.E)

P.E = 2 K.E

K.E = \frac{P.E}{2} = \frac{360}{2} = 180 KeV