Electric field at surface of lead-208 nucleus

AI Thread Summary
The discussion focuses on deriving the relationship between the radius of a lead-208 nucleus and the radius of a proton. The equation used is based on the volumes of the nucleus and proton, expressed as V = 208 Vp. The radius of the nucleus, r, is calculated using the formula r = (208)^(1/3)(1.20 × 10^-15 m). This relationship clarifies how the volume of the nucleus relates to the volume of protons. The participants express gratitude for the explanation, indicating a clear understanding of the derivation.
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1674151948824.png

The solution is,
1674151982159.png

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

Many thanks!
 
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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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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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