Calculating the radius of a nucleus of a Pd-118 atom

[Lt.Saccharine]

Here is the question I am attempting to solve :

A proton or neutron has a radius r of roughly 110-15 m, and a nucleus is a tightly packed collection of nucleons. Therefore the volume of the nucleus, (4/3)R3, is approximately equal to the volume of one nucleon, (4/3)r3, times the number N of nucleons in the nucleus: (4/3)R3 = N(4/3)r3. So the radius R of a nucleus is about N1/3 times the radius r of one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.310-15 m)N1/3. What is the radius of a palladium nucleus

So, in order to arrive at an answer I listed variables ,constants and equations that were known:
4/3*pi*R^3=4/3*pi*r^3*N
R= (N^1/3)*r
N=(1.3e-15)* (N^1/3)
r=1.0e-15


and so, in order decrease the number of variables given , I plugged in variables into variables:

R=(N^1/3)*r ---> R/r =(N^1/3)
N=(1.3e-15) * (N^1/3) ---> N=(1.3e-15)*R/r

so I substituted the new expression for N into:

4/3*pi*r^3 =4/3*pi*R^3 * N ----> 4/3*pi*r^3= 4/3*pi*R^3*((1.3e-15)*R/r):

pi and 4/3 cancel , show I should be left with:

r^4/(1.3e-15) = R
since r = 1e-15 , then R should equal :

(1e-15)^4/(1.3e-15)=R

And my answer is no where near the actually answer : Can you tell me what I am doing wrong

 

[Andrew Mason]

Here is the question I am attempting to solve :

A proton or neutron has a radius r of roughly 110-15 m, and a nucleus is a tightly packed collection of nucleons. Therefore the volume of the nucleus, (4/3)R3, is approximately equal to the volume of one nucleon, (4/3)r3, times the number N of nucleons in the nucleus: (4/3)R3 = N(4/3)r3. So the radius R of a nucleus is about N1/3 times the radius r of one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.310-15 m)N1/3. What is the radius of a palladium nucleus

So wouldn't the radius of the Pd-118 would be 1.31e-15 * N^{1/3} where N is the number of nucleons in the Pd-118 nucleus?

AM