Maximum Energy transfer in Compton Effect

[PhatPartie]

A photon having E = 14.7 keV energy scatters from a free electron inside a metal. What is the maximum energy the electron can gain from the photon?


Lamda'-lamda= change in lamda = h/mc (1-cos(theta)) ...having problems figuring out theta ... I know E=hc/lamda so lamda = hc/E ...so 1240eV/1.47*10^3 ev = .0843537 ...am i on the right track and if so where do i go from here? thanks

 

[Gamma]

Find $$E_0 - E_1$$ in terms of $$cos \theta$$ using compton equation.

See for what value of $$\theta$$, $$E_0 - E_1$$ is maximum.

$$(E_0 - E_1)_{max}$$ is the maximum energy imparted to the electron.

 

[thepatientmental]

Yes, you are on the right track. To find the maximum energy gained by the electron in the Compton effect, you need to use the equation: E' = E/(1 + (E/mc^2)(1 - cos(theta))), where E is the energy of the incident photon, E' is the energy of the scattered photon, m is the mass of the electron, and c is the speed of light.

In this case, the incident photon has an energy of 14.7 keV, so E = 14.7 keV. The mass of the electron is approximately 9.11 x 10^-31 kg and the speed of light is 3 x 10^8 m/s.

To find the value of theta, you can use the equation you mentioned, lambda' - lambda = h/mc (1-cos(theta)). Rearranging this equation, we get cos(theta) = 1 - (lambda'/lambda). Plugging in the values for lambda' and lambda, we get cos(theta) = 1 - (1.47 x 10^-10 m/1.24 x 10^-9 m) = 0.8823. Taking the inverse cosine of this value, we get theta = 30.4 degrees.

Finally, plugging in all the values in the equation for maximum energy transfer, we get E' = 14.7 keV/(1 + (14.7 keV/(9.11 x 10^-31 kg x (3 x 10^8 m/s)^2))(1 - cos(30.4 degrees))) = 14.7 keV/(1 + 5.93 x 10^-10)(1 - 0.8823) = 10.7 keV.

Therefore, the maximum energy the electron can gain from the photon is 10.7 keV. This shows that in the Compton effect, the maximum energy transfer occurs when the scattered photon is at a 90 degree angle from the incident photon.