# DeBroglie wavelength / particle phys

1. Jan 5, 2008

### opous

Would anybody be able to advise how I would approach the following question?

I know the deBroglie wavelength is h/p, but I'm unsure how to calculate p based on the kinetic/total energy...

2. Jan 5, 2008

### opous

OK, I've found a couple of different formulae which seem to hint at a way into this question:

$$\lambda = \frac{hc}{\sqrt{2mc^{2}K}}$$

where $$mc^{2}$$ is the rest mass energy and K is the Kinetic Energy.

This would mean that:

1. $$\lambda_{kaon} = \frac{1240eVnm}{\sqrt{2m_{kaon}c^{2}2.0GeV}}$$

where $$m_{kaon} = 493.7 MeV/c^{2}$$

Which I get to be $$\lambda = 0.88fm$$

However I'm not sure I'm on the right track here... I'm conscious that the question may require specific formulae for particle physics, and I have no idea where a formulae using the total energy can be found for this question..

Edit... in fact I'm virtually certain this is wrong, the text makes it clear that this is for LOW energy particles, not high. Back to square 1.

Last edited: Jan 5, 2008
3. Jan 5, 2008

### hage567

Start with the energy-momentum relationship $$E^2 = p^2c^2 + (m_0c^2)^2$$ where pc is the momentum term, $$m_0c^2$$ is the rest energy, and E is the total energy. You also know that the total energy is the sum of the kinetic energy and the rest energy. With some algebra, you can use these to find pc in terms of kinetic energy.

4. Jan 6, 2008

### opous

^ ok, so considering the total energy:

$$E_{tot} = pc + (m_{0}c^{2})$$

ie, $$pc = E_{tot} - m_{0}c^{2}$$

So I should be able to use:
$$\lambda = \frac{hc}{pc} = \frac{hc}{E_{tot} - m_{0}c^{2}}$$

for the total energy expression?

This would give $$\lambda_{muon} = 4.21fm$$ where $$m_{0,muon} = 105.7MeV/c^{2}$$

However, I still can't see a different way in calculating $$\lambda_{kaon}$$ than I have in my post above...

Last edited: Jan 6, 2008
5. Jan 6, 2008

### opous

Unless... I use this:

$$E^{2} = p^{2}c^{2} + m_{0}^{2}c^{4}$$
$$E = mc^{2} = KE + m_{0}c^{2}$$
$$p^{2}c^{2} = KE^{2} + 2KEm_{0}c^{2} + m_{0}^{2}c^{4} - m_{0}^{2}c^{4}$$
$$pc = \sqrt{KE^{2} + 2KEm_{0}c^{2}$$

So...

$$\lambda_{kaon} = \frac{hc}{\sqrt{KE^{2} + 2KEm_{0}c^{2}}} = 0.51fm$$

$$\lambda_{muon} = \frac{hc}{E_{tot} - m_{0}c^{2}} = 4.21fm$$

In which case $$\lambda_{muon} > \lambda_{kaon}$$

Does this look right? I'm a bit concerned I've done something totally wrong here!!

6. Jan 6, 2008

### hage567

You somehow lost the squares on the terms in the energy-momentum equation in your first steps when you worked out your expression for the muon.
I think your equation for the kaon is correct, so just fix the muon one.

7. Jan 6, 2008

### Staff: Mentor

This does not follow from the equation that hage567 gave you. In effect, you're trying to go from $c^2 = a^2 + b^2$ to $c = a + b$ which doesn't work because $\sqrt{a^2 + b^2} \ne a + b$.

8. Jan 7, 2008

### opous

Ah, got you, so redoing the Muon one:

$$\lambda_{muon} = \frac{hc}{pc} = \frac{hc}{\sqrt{E_{tot}^{2} - (m_{0}c^{2})^{2}}}$$

Giving

$$\lambda_{muon} = 3.21 fm$$

I think I get that now, thanks for the clarification hage567 and jtbell!