Probing Nuclei with Electron Scattering

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SUMMARY

The discussion focuses on calculating the acceleration of an electron in the electric field of a lead nucleus, specifically at distances of nR, R, and R/n from the nucleus center. The charge of the lead nucleus is +82e, and the relevant equations include F=qE, E=1/4πϵ₀ * q/r², and a=F/m. The user successfully completed calculations for distances nR and R but encountered difficulties with R/n, highlighting the importance of applying Gauss' law to understand electric fields within the nucleus.

PREREQUISITES
  • Understanding of electric fields and forces (F=qE)
  • Familiarity with Gauss' law and its applications
  • Knowledge of basic electromagnetism concepts
  • Proficiency in calculus for solving physics problems
NEXT STEPS
  • Study Gauss' law and its implications for electric fields within spherical charge distributions
  • Explore the concept of electric field strength within a nucleus
  • Review the derivation of electric field equations for point charges and spherical charge distributions
  • Practice problems involving the acceleration of charged particles in electric fields
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and nuclear physics, as well as anyone interested in the behavior of charged particles in electric fields.

TomCass
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Homework Statement


To study the structure of the lead nucleus, electrons are fired at a lead target. Some of the electrons actually enter the nuclei of the target, and the deflection of these electrons is measured. The deflection is caused by the charge of the nucleus, which is distributed approximately uniformly over the spherical volume of the nucleus. A lead nucleus has a charge of +82e and a radius of R.

(Successfully completed) A. Find the acceleration of an electron at a distance of n R from the center of a lead nucleus.
Use ϵ_0 for the permittivity of free space, e for the magnitude of the charge on an electron, and m_e for the mass of an electron.

(Successfully completed) B. Find the acceleration of an electron at a distance of R from the center of a lead nucleus.

(Stumped, for some reason) C. Find the acceleration of an electron at a distance of R/ n from the center of a lead nucleus.

Homework Equations


F=qE

E=1/4pi(epsilon_naught)*q/r^2

a=F/m

The Attempt at a Solution


For example, this is what the solution to A is: [82(e)^2]/4(pi)(epsilon_naught)(m_e)(nR)^2

It seems to me that C should be the exact same process for A except with (R/n) instead of (nR). This is not correct. I was wondering if someone had a hint as to whether or not R now being divided would have an impact on an earlier calculation.

Any help is greatly appreciated! Thanks a bunch! And if I didn't clarify something enough in the explanation, feel free to ask!
 
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TomCass said:

Homework Statement


To study the structure of the lead nucleus, electrons are fired at a lead target. Some of the electrons actually enter the nuclei of the target, and the deflection of these electrons is measured. The deflection is caused by the charge of the nucleus, which is distributed approximately uniformly over the spherical volume of the nucleus. A lead nucleus has a charge of +82e and a radius of R.

(Successfully completed) A. Find the acceleration of an electron at a distance of n R from the center of a lead nucleus.
Use ϵ_0 for the permittivity of free space, e for the magnitude of the charge on an electron, and m_e for the mass of an electron.

(Successfully completed) B. Find the acceleration of an electron at a distance of R from the center of a lead nucleus.

(Stumped, for some reason) C. Find the acceleration of an electron at a distance of R/ n from the center of a lead nucleus.

Homework Equations


F=qE

E=1/4pi(epsilon_naught)*q/r^2

a=F/m

The Attempt at a Solution


For example, this is what the solution to A is: [82(e)^2]/4(pi)(epsilon_naught)(m_e)(nR)^2

It seems to me that C should be the exact same process for A except with (R/n) instead of (nR). This is not correct. I was wondering if someone had a hint as to whether or not R now being divided would have an impact on an earlier calculation.

Any help is greatly appreciated! Thanks a bunch! And if I didn't clarify something enough in the explanation, feel free to ask!

It makes a huge difference within the boundary of the nucleus. What does Gauss' law say exactly?

Have you done problems dealing with the gravitational field strength *within* the earth? Same principle.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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