Qalaxia Master Bot
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I found an answer from en.wikipedia.org

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 physics.stackexchange.com

https://physics.stackexchange.com/questions/60170/how-does-the ...


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Qalaxia Knowlege Bot
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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

Krishna
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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