Electric field at surface of lead-208 nucleus

In summary, the given conversation discusses the relationship between the volume of a nucleus and the volume of a proton, represented by the respective radii. By equating the two volumes, the expression ## r = (208)^{1/3}(1.20 \times 10^{-15} m)## is obtained, which shows the proportional relationship between the two radii.
  • #1
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
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The solution is,
1674151982159.png

However why is ## r = (208)^{1/3}(1.20 \times 10^{-15} m)##

Many thanks!
 
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  • #2
Just write the relationship given, between the volume of proton and volume of nucleus in terms of the respective radius. (##V=208 V_p##).
 
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  • #3
nasu said:
Just write the relationship given, between the volume of proton and volume of nucleus in terms of the respective radius. (##V=208 V_p##).
Thanks for your reply @nasu!

I will use ##r## for the radius of the nucleus and ##R## for the radius of the proton.

## \frac {4{\pi}r^3} {3} = 208\frac {4{\pi}R^3} {3}##
## r =(208R^3)^{1/3}##
## r = 208^{1/3}R ##
## r = (208)^{1/3}(1.20 \times 10^{-15} ## ##m)##

Thank you I see how they got it now!
 
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FAQ: Electric field at surface of lead-208 nucleus

What is the electric field at the surface of a lead-208 nucleus?

The electric field at the surface of a lead-208 nucleus can be calculated using Coulomb's law. Given that lead-208 has 82 protons, the electric field is determined by the charge of these protons and the radius of the nucleus. The electric field is approximately 1.3 x 1021 N/C.

How do you calculate the electric field at the surface of a lead-208 nucleus?

To calculate the electric field, you use the formula E = k * (Q / r2), where k is Coulomb's constant (8.99 x 109 N m2/C2), Q is the total charge (82 protons, or 82 * 1.6 x 10-19 C), and r is the radius of the nucleus (approximately 7.1 x 10-15 m).

Why is the electric field at the surface of a lead-208 nucleus significant?

The electric field at the surface of a lead-208 nucleus is significant because it provides insights into the electrostatic forces that act on protons within the nucleus. These forces play a crucial role in nuclear stability and reactions, influencing phenomena such as alpha decay.

What factors affect the electric field at the surface of a lead-208 nucleus?

The primary factors that affect the electric field at the surface of a lead-208 nucleus are the number of protons (which determines the total charge) and the radius of the nucleus. Any changes in these parameters will alter the electric field strength.

Can the electric field at the surface of a lead-208 nucleus be directly measured?

Directly measuring the electric field at the surface of a lead-208 nucleus is extremely challenging due to the tiny scale and the high field strength. Instead, it is typically calculated using theoretical models and experimental data on nuclear properties.

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