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

 

[kyle8921]

?

I appreciate your attempt at solving the problem using various equations and variables. However, there are a few things that need to be clarified.

Firstly, the equation you have used, (4/3)R^3 = N(4/3)r^3, is not correct. This equation assumes that the volume of a nucleus is equal to the volume of one nucleon multiplied by the total number of nucleons, which is not accurate.

Secondly, the value of r that you have used, 1.0e-15, is not the correct radius of a proton or neutron. The actual value is approximately 0.8e-15 m.

Lastly, the equation N = (1.3e-15) * (N^1/3) does not make sense mathematically. It is important to use valid equations when solving scientific problems.

Instead, the correct equation to use for calculating the radius of a nucleus is R = (1.3e-15 m) * N^(1/3). This equation takes into account the fact that the radius of a nucleus is proportional to the number of nucleons it contains.

Therefore, in order to calculate the radius of a Pd-118 nucleus, we need to know the number of nucleons in this nucleus. Since Pd-118 has an atomic number of 46, we can assume that it has 118 nucleons (46 protons and 72 neutrons).

Using the equation R = (1.3e-15 m) * N^(1/3), we get R = (1.3e-15 m) * (118)^(1/3) = 2.8e-15 m. This is the correct answer for the radius of a Pd-118 nucleus.

In summary, it is important to use accurate equations and values when solving scientific problems. I hope this explanation helps you understand where you went wrong and how to arrive at the correct answer. Keep up the good work!