# Atomic Nucleus

1. Feb 29, 2012

### glebovg

1. The problem statement, all variables and given/known data

In order to study the atomic nucleus, we would like to observe the diffraction of particles whose de Broglie wavelength is about the same size as the nuclear diameter, ~14 fm for a heavy nucleus such as lead. What kinetic energy should we use if the diffracted particles are

a) electrons
b) neutrons
c) alpha particles (m = 4 u)

2. Relevant equations

$p = \frac{h}{\lambda}$
$E^{2}=(pc)^{2}+(mc^{2})^{2}$

3. The attempt at a solution

Am I supposed to use these equations for each part?

$pc = \frac{hc}{\lambda}$
$E^{2}=(pc)^{2}+(mc^{2})^{2}$
$K=E-mc^{2}$

However, for c) $E \approx mc^{2}$ so we should use a non-relativistic formula $K=\frac{p^{2}}{2m}=\frac{(pc)^{2}}{2mc^2}$ instead.

Last edited: Feb 29, 2012
2. Mar 1, 2012

### gomboc

So, you're firing various particles at a 14 fm lead nucleus to measure their diffraction.

We know that in order to make any measurement in physics, the wavelength of the photon/particle taking the measurement must be similar to, or even better, smaller than the dimension you wish to measure - in this case, 14 fm.

You're simply using the relativistic energy equation to determine what energy is needed to give the measuring particle (electrons, neutrons, alpha particles) a de Broglie wavelength small enough to take a decent measurement.

And for the alpha particles, yes, you are correct, you would obtain the correct answer using the non-relativistic formula. It's a teeny bit less precise than doing it the same way as the other particles, but the significant figures will probably end up giving the same kinetic energy result.

Last edited: Mar 1, 2012
3. Mar 1, 2012

### glebovg

So is my approach correct? Is there another way of doing this? For some reason my instructor wanted us to use a different formula. My instructor says we are supposed to use the formula for V. I am not sure which formula she is referring to. Perhaps, we need to use

$qV=\frac{1}{2}mv^{2}$ and $p=mv$, which give $V=\frac{h^{2}}{2mq\lambda^{2}}$.

4. Mar 1, 2012

### glebovg

Can anyone help?

5. Mar 1, 2012

### ehild

Your approach is correct. qV=1/2 mv^2 applies only charged and non-relativistic particles.

ehild

6. Mar 1, 2012

### glebovg

That is exactly what I thought. What do you suppose V is? Is there another way of doing this? Using the formula for V?

Last edited: Mar 1, 2012
7. Mar 2, 2012

### ehild

V can be the accelerating voltage, but it does not apply for neutrons.
Or v can be the velocity and your professor might want you to use formulae KE= mc^2/Sqrt(1-(v/c)^2)-mc^2 and p=mv/Sqrt(1-(v/c)^2).

ehild

8. Mar 2, 2012

### glebovg

Apparently we were supposed to use $V=\frac{h^{2}}{2mq\lambda^{2}}$ for kinetic energy. However, voltage is measured in volts, or joules per coulomb, whereas energy is measured in joules, how is it possible?

9. Mar 2, 2012

### ehild

The energy of an electron is often given in electron-Volts (eV). In case of KE, the value of voltage is given which would accelerate the charged particle to that KE. The kinetic energy is than the voltage multiplied by the charge. If it is said that the energy of an electron is 10 MeV, it means 107 V * 1.6x10-19 C =1.6 x10-12J.

For non-relativistic velocities, KE=p2/(2m). p=h/λ. KE=qV. So

V=KE/q=p2/(2mq)=h2/(2mqλ2).

ehild

10. Mar 2, 2012

### glebovg

So which approach is correct? Would we be able to use the formula for V in b) and c)?

11. Mar 3, 2012

### ehild

If you compare pc and mc2 for the particles you will see that pc>>mc2 for the electron, but mc2>>pc for the other two particles.

You can write the relativistic formula E2=(pc)2+(mc2)2 in the form

$$(pc)^2=E^2-(mc^2)^2=(E-mc^2)(E+mc^2)=K(E+mc^2)$$

that is,

$$K=\frac{(pc)^2}{E+mc^2}$$.

This formula is valid for all particles. Calculate the kinetic energy. You might convert it to eV (electron-volt units) by dividing by 1.6X10-19.

E≈mc^2 for the heavy particles, so K≈p^2/(2m), the classical approach can be applied when you answer b) and c). But the voltage formula gives voltage, not energy, and it has no sense for the neutrons.