Checking Chadwick's statement about the mysterious neutral radiation ?

[jeebs]

Checking Chadwick's statement about the mysterious "neutral radiation"?

Hi,
I have this interesting little problem about Chadwick's identification of the neutron. As the story goes, Curie and Joliot were firing poloniom-sourced alpha particles at beryllium, causing the emission of a neutral, penetrating radiation. They noticed that it could eject protons from hydrogen containing material, and decided it was gamma radiation causing the proton ejections via a sort of Compton-style effect. Chadwick apparently did a calculation where he showed that the incident photons would have to have an energy of 50MeV to eject protons whose top speed was 3*10⁹ cm s^-1. Since this is very high energy for a photon, he proposed it was actually neutrons that were being observed.

I am supposed to be verifying his statement that the incident photons would have to be at least 50MeV if they were indeed photons. However, I am getting weird answers, the closest I have gotten is 117MeV. What I'll do here is explain the bits I've worked out that I am pretty sure of...

If I'm not mistaken, the standard Compton effect involves a photon of wavelength $$\lambda$$ coming in, interacting with the atomic electron, deflecting through an angle x, transferring momentum to the electron and exiting with a longer wavelength $$\lambda'$$. The equation is $$\lambda' - \lambda = \frac{h}{m_ec}(1-cos(x))$$.
As it is a proton that is being interacted with in this problem, I am substituting m_p for m_e.

I also thought that the maximum transfer of momentum from photon to proton will occur when the deflection angle is 180 degrees. This, I was thinking, should allow us to achieve the proton's acceleration to 0.1c with the lowest possible energy photon.

So, setting x=180, I can say that
$$\lambda' - \lambda = 2\frac{h}{m_pc} = \frac{2*6.63*10^-^3^4Js}{1.67*10^-^2^7Kg*3*10^8c} = 2.65*10^-^1^5m.$$

In other words, the difference in wavelength before and after interaction is 2.65*10^-15m. However, using de Broglie's $$E= hf = hc/\lambda$$ we can say that this corresponds to an energy transfer of $$\Delta E= hf = hc/\lambda - hc/\lambda'$$ ie. $$\frac{1}{\Delta E} = \frac{\lambda -\lambda'}{hc} = \frac{2.65*10^-^1^5m}{6.63*10^-^3^4*3*10^8} = 1.33*10^1^0 J^-1$$
In other words $$\Delta E = 7.5*10^-11J = 469MeV$$. This is clearly wrong when you see what I found next.

I have then worked out the proton's relativistic momentum via
$$p_p' = \gamma m_pv_p = \frac{m_pv_p}{\sqrt{1-\frac{v_p^2}{c^2}}} = \frac{1.67*10^-^2^7*3*10^7ms^-^1}{\sqrt{1-0.1^2}} = 5.03*10^-^2^0 Kgms^-^1 = 94.3MeV/c$$

Momentum must be conserved, so that if we denote the quantities after the deflection with a ' we get $$p_\gamma = p_\gamma' + p_p'$$. I know what p_p' is and I need $$p_\gamma$$ to use in $$E_\gamma = p_\gamma c$$.What I'm thinking is that the reflected photon needs to come away with a finite amount of momentum since the proton cannot take the 469MeV that my Compton scattering equation says it would. How you find this i do not know...

I have no idea how to proceed with this. Anyone got any suggestions?
Thanks.

 

[quenderin]

Well $$\Delta E = hc/ \lambda - hc/ \lambda '$$ sounds right, but I don't think the next step follows!

 

[jeebs]

how does it not follow? the proton ejection speed is definitely 0.1c. It's been a while since I calculated or derived relativistic momentum so I'm using that equation a bit blindly, but assuming it's right we get a proton energy gain that is far less than the photon energy loss. Where's the energy went?

 

[quenderin]

heh it's just an algebraic error when you took $$1/ \Delta E$$:

$$\frac{1}{hc/\lambda - hc/\lambda '} \neq \frac{\lambda - \lambda'}{hc}$$

 

[jeebs]

ahh, what a schoolboy error...