Probing Atomic Nuclei with Electron Scattering.

In summary, the electron's acceleration at a distance of 3.00 R from the center of a lead nucleus is -9.9*10^-20.
  • #1
stylez03
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0

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 = 7.10×10^−15

Find the acceleration of an electron at a distance of 3.00 R from the center of a lead nucleus.

Homework Equations



Charge of Electron = -1.60*10^-19

Mass of Electron = 9.11*10^-31

[tex] F = \frac {k*q*Q} {r^{2}} [/tex]

[tex] k = \frac {1} {4*pi*e_o} [/tex]

[tex] e_o = 8.85*10^{-12} [/tex]

[tex] A = \frac {F} {m} [/tex]

The Attempt at a Solution



=> [tex] F = \frac {\frac {1} {4*pi*8.85*10^{-12}}*-1.60*10^{-19}*82*1.60*10^{-19}} {(3*(7.10*10^{-15})^{2}} [/tex]
 
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  • #2
Looks good so far.
 
  • #3
berkeman said:
Looks good so far.

When I get down to a numeric answer, it doesn't seem right.

[tex] \frac {(6.95*10^{-12}*(-2.17*10^{-36}} {1.51*10^{-28}} [/tex]

= [tex] -9.9*10^{-20} [/tex]
 
  • #4
First, remember that you are solving for a=F/m. Second, the first term in the numerator went from 1/10^-12 to 10^-12. Maybe a misplaced "-" sign?
 
  • #5
berkeman said:
First, remember that you are solving for a=F/m. Second, the first term in the numerator went from 1/10^-12 to 10^-12. Maybe a misplaced "-" sign?

Yes you're right the first term was actually incorrect:

[tex] \frac {(8.99*10^{9})*(-2.17*10^{-36})} {1.51*10^{-28}} [/tex]

F = 1.00063

[tex] A = \frac {F} {M} [/tex]

[tex] A = \frac {1.00063} {9.11*10^{-31}} [/tex]

[tex] A = 1.09*10^{30} [/tex]

Can you confirm this numeric answer? I think it should be 10^32?

I solved for the case when it's just distance of "R" and it came out to be 10^32, so shouldn't these other distances have a factor of the same 10^32? I have limited of tries and I just wanted to be sure I wasn't doing the math incorrectly and making a careless mistake. Thank you
 
  • #6
I get a different denominator number for F, but I could just be messing up the calculator (I had to borrow one). I have to bail for a while -- just re-check your numbers. One trick is to punch the numbers into the calculator in a different order to help you catch any calculation errors. Good luck.
 

1. What is electron scattering and how is it used to probe atomic nuclei?

Electron scattering is a technique in which a beam of high-energy electrons is directed at a target material, such as an atomic nucleus. As the electrons interact with the target, they can probe the structure and properties of the nucleus. This is possible because electrons have a much smaller wavelength compared to other particles, allowing them to interact with the nucleus at a subatomic level.

2. How does the energy of the electron beam affect the results of electron scattering experiments?

The energy of the electron beam is a crucial factor in electron scattering experiments. Higher energy electrons have a shorter wavelength, which allows for more precise measurements of the nucleus. Additionally, higher energy electrons have a greater penetrating power, allowing them to pass through thicker target materials and probe deeper into the nucleus.

3. What information can be obtained from electron scattering experiments?

Electron scattering experiments can provide valuable information about the structure and properties of atomic nuclei. This includes the size, shape, and distribution of protons and neutrons within the nucleus, as well as the nuclear forces that hold these particles together.

4. How does electron scattering compare to other methods of studying atomic nuclei, such as particle accelerators?

Electron scattering offers several advantages over other methods of studying atomic nuclei. It is a non-destructive technique, meaning that the target material is not altered or destroyed during the experiment. It also allows for precise measurements of the nucleus without the need for expensive equipment, making it a cost-effective option for studying nuclear structure.

5. What are the practical applications of studying atomic nuclei through electron scattering?

Studying atomic nuclei through electron scattering has a wide range of practical applications. It can help us understand the properties of nuclear matter and the behavior of particles within the nucleus, which is crucial for fields such as nuclear energy and medicine. Electron scattering can also aid in the development of new technologies, such as particle accelerators and nuclear weapons detection methods.

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