# Homework Help: Nuclear Physics - HELP!

1. Aug 10, 2012

### Ruro

Hello, I'm an AS student and probably going to study A2. I was wondering if someone tell me how to work out these questions, step by step?

Relationship between Radius of Nucleus And Mass Number
1. By assuming that the density of the nucleus is the same for all elements in the periodic table, show that the radius of the nucleus of any element satisfies

R α A^(1/3)

α = directly proportional

where A is the mass number of the nucleus. Assume nucleus is a spherical (4/3)π(r)^3

Determining the approximate size of the Nucleus
Rutherford was able to estimate the size of the nucleus by firing a stream of alpha particles at a thin layer of gold leaf a few atoms thick. Consider an alpha particle is traveling towards the Gold nucleas. The alpha particle is positively charged and the Gold nucleus is postively charged. Therefore, for head on collisions between the alpha and the nucleus, there will be a time when all of the Kinetic energy of the alpha particles is converted into electrical potential energy (the alpha particle gets slower and slower as it gets closer to the nucleus before momentarily stopping, then it gets repelled back in the same direction that it came from. This is similar to when you throw a ball into the air. The balls velocity decreases as it travels upwards until it momentarily stops at the point when all the kinetic energy has been converted into gravitational potential energy. The ball then falls back to the ground).

It is possible to determine the distance between the nucleus and the alpha particle at the point where all the kinetic energy has been converted into electrical potential energy. This distance provides an estimate for size of the nucleus. Obviously, the fast the initial velocity of the alpha particle, the closer the alpha particle will get tothe nucleus and the more accurate the estimated size of the nucleus will be.

2. Your task is to derive a general expression for the radius of a nuclus of atomic number Z in terms of the initial velocity of the alpha particle. You must then estimate the size of the nucleus by assuming that the initial velocity of the alpha particle is 2 x 10^7 m/s. Hint: An expression for the electrical potential energy VE can be inferred from the relation:

FE = - (dVE / dr)

where FE is the Coulomb Force acting between two charges Q1 and Q2 seperated by a distance r:

FE = (1/4π weird symbol) x (Q1Q2 / r^2)

Electrical potential energy = (1/4πweirdsymbol) x (Q1Q2/r)

NOTE: Weird symbol looks like a euro sign and has a small zero next to it. I'm not sure what it is. c:

1. I'm confused as to how you show that the radius of the nucleus of any element is satisfied. I used Gold, and I did:

Mass number of Gold = 197

R α 197^(1/3)
R α 5.819

After that, I'm a little lost. Do I do the following?

(4/3) x π x 5.819^3 = 825.19

That seems wrong.

2. This one, I'm just very lost. I don't know where to start!

Last edited by a moderator: Aug 10, 2012
2. Aug 10, 2012

### cepheid

Staff Emeritus
EDIT: welcome to PF!

The density is the same for all elements. So density (rho) = mass/volume = A/(4/3 pi * R^3)

(4/3 pi R^3) = A/rho

R^3 = (3/4)A/(rho*pi)

R = (a bunch of constants)* A^(1/3)

R is proportional to the cube root of A.

Basically, the dimensional argument is that if the density is constant, the volume has to scale linearly with mass, which means that the cube of the radius has to scale linearly with mass.

3. Aug 11, 2012

### Ruro

I completely forgot about the whole Density = Mass/ Volume. Thank you again for your help! It makes more sense now.

So what you're basically saying is that to show that the radius of the nucleus of any element satisfies R α A^(1/3), I need to rearrange the equations with some of the formulae given instead of using a real element mass number as an example? Because if I try that, I don't know how to find the density of the elements.

If someone could help me with Question 2, I'd very much appreciate it.

4. Aug 13, 2012

### Ruro

Bumping This!

Can someone help? Please?

5. Aug 13, 2012

### cepheid

Staff Emeritus
Well...yeah. I mean if you're asked to show that a scaling relation between two variables applies, isn't the logical step to manipulate the equations to express one of the variable in terms of the other, and see if that relation is what comes up?

How would you use a real mass number as an example? All you know is that R and A are proportional in some way. That does not give you enough information to figure out R just by knowing A.

Let me give you another example. We can find out from experiment that the acceleration 'a' of an object is directly proportional to the net force 'F' applied to it. In other words, the acceleration varies linearly with force. What this means is that the acceleration is equal to some constant times the force. This constant is called the constant of proportionality (It's the slope in the linear relation). But if we don't know what that constant IS, then we can't compute the acceleration given the force, because we don't have a full equation relating one to the other, we just have a proportionality. $a \propto F$ means that if I double the force, I'll double the acceleration. If I triple the force, I'll triple the acceleration. If I make the force one quarter of what it was before, the acceleration will decrease by a factor of four as well. That's what direct proportionality *means.* But, in this example, if you tell me that F = 2 N, that doesn't give me enough info to compute a, because all I know is that a is equal to *some constant* times F, but I don't know what that constant is. (Of course, in *reality* we know from hundreds of years of experiment that the constant of proportionality between acceleration and force is the inverse of the mass i.e. a = (1/m)*F, that's just Newton's 2nd Law. But I'm talking about a hypothetical situation where you don't know this constant (slope) of the linear relation, all you are told is that it is linear).

Similarly in your example, the scaling relation tells you that if you increase the mass by a factor of 8, the radius will double, and if you increase it by a factor of 27, the radius will triple, et cetera. But you don't know the constant of proportionality C in the equation R = C*A1/3, then you can't compute R for a given value of A. So I don't understand what it is that you were trying to compute in your attempt above.

The question doesn't ask you to try to compute the density of the elements. All it tells you to do is to assume that, whatever it is, it's the same for all of them.

Okay, so you have an expression for the potential energy of the system as the one positive charge (the alpha particle) approaches the other positive charge (the atomic nucleus). And as you can see, it depend only on r, which is the separation (distance) between the two charges. So as this separation decreases (r gets smaller), what happens to the potential energy of the system? What does that imply about what is happening to the the kinetic energy of the alpha particle? (Remember that energy is conserved). So, from this fact, how would you compute the value of r at which the the alpha particle would be stopped?

The "weird symbol" is the Greek letter epsilon: ε (you can find it in the "quick symbols" section to the right of the reply box). Epsilon is the symbol for electric permittivity. In this case, the epsilon has a subscript of 0: ε0. The subscript of 0 implies that this is referring to the electric permittivity of free space, rather than the electric permittivity inside some material. I'm surprised you haven't come across this in studying electromagnetism:

http://en.wikipedia.org/wiki/Vacuum_permittivity

You can produce subscripts and superscripts in your posts by using the X2 and X2 buttons above the reply box, respectively. Note that when a quantity has a subscript 0, we usually add "naught" (pronounced "not") when we say its name. Naught is just another word for 0 or "nothing." So ε0 would be called "epsilon-naught." Epsilon-naught is the symbol for the electric permittivity of free space.