Proton Velocity (potential/field/charge)

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SUMMARY

The discussion focuses on calculating the minimum speed required for two protons to undergo nuclear fusion, specifically addressing the conditions under which their surfaces come into contact. The relevant equations include the potential energy equation \( qV = \frac{1}{2}mv^2 \) and the electric potential \( V = \frac{kq}{r} \). The calculated minimum speed for fusion is determined to be approximately \( 7.58 \times 10^6 \) m/s, contrasting with an initial incorrect calculation of \( 1.5 \times 10^7 \) m/s. Additionally, the discussion raises questions about the distribution of kinetic energy among the particles involved.

PREREQUISITES
  • Understanding of nuclear fusion processes
  • Familiarity with electric potential and fields
  • Basic knowledge of kinetic energy equations
  • Proficiency in algebra and physics problem-solving
NEXT STEPS
  • Study the principles of nuclear fusion in stellar environments
  • Learn about electric fields and potentials in particle physics
  • Explore the concept of kinetic energy distribution in colliding particles
  • Investigate the role of temperature and pressure in fusion reactions
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Students studying physics, particularly those focusing on nuclear physics, astrophysics, or anyone interested in the mechanics of particle collisions and fusion processes.

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Homework Statement


(background info)
The source of the sun's energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact.

Assume both protons are moving with the same speed and they collide head-on. If the radius of the proton is 1.2*10^-15m, what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is 1.67*10^-27kg.


Homework Equations





The Attempt at a Solution



qV=(1/2)mv^2
V=kq/r
kq^2/r=(1/2)mv^2
plugging in for the numbers gives me 1.5*10^7

The answer should be 7.58*10^6
 
Physics news on Phys.org
How close will the proton's centers be to each other when their surfaces just meet?

How many particles will share the required kinetic energy? How will it be distributed between them?
 

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