- #1
jeebs
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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 mp for me.
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 pp' 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.
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 mp for me.
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 pp' 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.
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