Problem Description: A proton is essentially a 2.4 fm diameter sphere of charge

Question

Problem Description:
A proton is essentially a 2.4 fm diameter sphere of charge andfusion occurs only if two protons come into contact with eachother. This requires extrarodinarily high temperatures due to thestrong repulsion between the protons. Recall that the averagekinetic energy of a gas particle is 3/2kbT.

a) Suppose two protons each with exactly the average kineticenergy, have a head on collision. What is the minimumtemperature for fusion to occur?

My Question:
Understanding that 1/2mv2 =3/2kbT I am unsurehow to use this relationship to solve for the min temperature. Do Ineed other relationships? If so what are they?

Explanation / Answer

Yeah, you need just a bit more... electrical potential energyto relate to the distance between the protons. . By conservation of energy... .     initial K of both protons = final potential energy .      2K     = kq q / r .       K = k q2/ 2r = 8.99 x 109 * (1.60 x10-19)2 / 2 * 2.4 x10-15 =  4.795 x 10-14 Joules . This is the kinetic energy of each proton, so now... .    K = (3/2) kT          T = 2 K / 3k     or .        T = 2 * 4.795 x10-14 / 3 * 1.38 x 10-23=    2.316 x109 Kelvin   is thetemperature         PLEASERATE, thanks! . Note: fusion occurs at the center of the sun at a temperaturethat is much lower than this... because most of the protons aremoving much slower, but just a few are moving much faster thanaverage and can actually "fuse". We wouldn't want ALL of theprotons (or even half) to fuse at the same time... that would bevery bad. . Note: fusion occurs at the center of the sun at a temperaturethat is much lower than this... because most of the protons aremoving much slower, but just a few are moving much faster thanaverage and can actually "fuse". We wouldn't want ALL of theprotons (or even half) to fuse at the same time... that would bevery bad.

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