Energy conservation in an alpha-scattering experiment

In summary, an alpha particle has a KE of 6 x 10^6 x e joules and a PE of kq1q2/d, where k=9 x 10^9, q1=4e, q2=Ze=79e. Plugging in these values gives an incorrect answer of 2 x 10^(-14) meters.
  • #1
Krushnaraj Pandya
Gold Member
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Homework Statement


In scattering experiment, find distance of closest approach if a 6 MeV alpha particle is used

2. The attempt at a solution
initially KE of alpha particle is 6 x 10^6 x e joules and 0 PE, finally its PE is kq1q2/d, k=9 x 10^9, q1=4e, q2=Ze=79e (assuming gold), d is distance of closest approach, e is charge on electron. Plugging in the values gives an incorrect answer. The correct answer (far from it) is 2 x 10^(-14) m. Is there a mistake I'm making?
 
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  • #2
Which answer did you get and can you show the steps?
At closest approach both the alpha particle and the gold nucleus will move a bit (conservation of momentum!) but this is a small effect.
 
  • #3
Your approach is correct and I get your “correct” answer. I see only 1 definite mistake, the charge of an alpha particle isn’t 4. However I suspect you big problem is units. You have k in SI units and PE in MeV and two charges in electron charges. I’m sure you converted some of them, but did you get all of them? Is your answer off by 10^19? If so, you missed one.
 
  • #4
Why am I not getting any alerts to replies to all my posts?
Anyway, I'll be back in a while and show my work
 
  • #5
Cutter Ketch said:
Your approach is correct and I get your “correct” answer. I see only 1 definite mistake, the charge of an alpha particle isn’t 4. However I suspect you big problem is units. You have k in SI units and PE in MeV and two charges in electron charges. I’m sure you converted some of them, but did you get all of them? Is your answer off by 10^19? If so, you missed one.
Right, its 2e. Thank you for pointing it out, I'll try that again
 
  • #6
mfb said:
Which answer did you get and can you show the steps?
At closest approach both the alpha particle and the gold nucleus will move a bit (conservation of momentum!) but this is a small effect.
After correcting the charge, I'll redo my calculations and post here within the next 24 hrs
 

Related to Energy conservation in an alpha-scattering experiment

1. What is energy conservation in an alpha-scattering experiment?

Energy conservation in an alpha-scattering experiment refers to the principle that energy is neither created nor destroyed, but rather transferred from one form to another. In this experiment, the energy of the incoming alpha particle is conserved as it interacts with the nucleus of the target atom, resulting in changes in the direction and energy of the alpha particle.

2. How is energy conservation demonstrated in an alpha-scattering experiment?

Energy conservation can be demonstrated by measuring the initial and final energies of the alpha particles in the experiment. The sum of the initial and final energies should be equal, indicating that energy has been conserved throughout the interaction between the alpha particle and the target atom nucleus.

3. Why is energy conservation important in an alpha-scattering experiment?

Energy conservation is important in an alpha-scattering experiment because it allows scientists to study the interactions between particles and atoms and understand the underlying physical processes. It also helps to validate the laws of conservation of energy and momentum, which are fundamental principles in physics.

4. What factors affect energy conservation in an alpha-scattering experiment?

The energy conservation in an alpha-scattering experiment can be affected by factors such as the mass and charge of the alpha particle and target atom, the distance between them, and the angle at which the alpha particle approaches the target atom nucleus. These factors can influence the amount of energy transferred and the resulting changes in direction and energy of the alpha particle.

5. How does energy conservation relate to the outcome of an alpha-scattering experiment?

The principle of energy conservation is essential in understanding the outcome of an alpha-scattering experiment. It helps to explain the changes in the direction and energy of the alpha particles as they interact with the target atom nucleus. By conserving energy, scientists can accurately predict and interpret the results of the experiment and gain insights into the structure and behavior of atoms.

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