Finding the potential difference of an alpha particle.

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SUMMARY

The discussion focuses on calculating the potential difference (\Delta V) required to accelerate an alpha particle to reach the surface of a 15 x 10-15 m nucleus of Uranium-238 (^{238}U). The kinetic energy (KE) of the alpha particle is determined using the equation KE = q\Delta V, where q represents the charge of the alpha particle. Participants emphasize the importance of conservation of energy, noting that while the kinetic energy is zero at the nucleus's surface, the potential energy (PE) is significant. The discussion concludes that knowing the charge of the alpha particle allows for the calculation of the necessary potential difference to achieve the required kinetic energy.

PREREQUISITES
  • Understanding of kinetic energy and potential energy concepts
  • Familiarity with the equations KE = q\Delta V and E = \gamma_{p}mc²
  • Knowledge of the properties of alpha particles and Uranium-238
  • Basic grasp of conservation of energy principles
NEXT STEPS
  • Calculate the charge and mass of an alpha particle
  • Explore the concept of electric potential and its relation to kinetic energy
  • Study the conservation of energy in nuclear physics contexts
  • Investigate the role of magnetic fields (B) in particle motion
USEFUL FOR

Students and professionals in nuclear physics, particle physicists, and anyone interested in the dynamics of alpha particles and their interactions with atomic nuclei.

elephantorz
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1. Through what potential difference ([tex]\Delta[/tex]V) would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15 x 10^(-15) m [tex]^{238}[/tex]U nucleus?
2. KE = q[tex]\Delta[/tex]V
E = [tex]\gamma_{p}[/tex]mc²
r = [tex]\frac{mv}{qB}[/tex]
3. Ok, I feel like I am close to what I really need but I am lacking something...
You can get the charge (q) and mass for both the particle and the U nucleus from the information given, you can also get the radius of the nucleus if needed.

I have that, I just seem to have two unknowns for every equation I am using, if I solve for u in E = [tex]\gamma_{p}[/tex]mc² I don't know Energy, and if I solve for v or B in
r = [tex]\frac{mv}{qB}[/tex] I don't know one of them, what am I missing?
 
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Where did the B come from?

If the alpha particle just reaches the surface of the nucleus, its KE is zero, but PE is not.

But how much KE it should have had to reach that point? Remember conservation of energy.

Knowing the charge of the alpha particle, what should be ∆V to give that KE at a point far away from the nucleus?

There, I just solved it for you.
 

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