Electric field at surface of lead-208 nucleus

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

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

Many thanks!

 

[nasu]

Just write the relationship given, between the volume of proton and volume of nucleus in terms of the respective radius. ($V=208 V_p$).

 

[ChiralSuperfields]

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!