Calculate the percentage change in photon energy during collision like that in Fig. 38-5 for$\mathit{\varphi }\mathbf{=}{\mathbf{90}}^{\mathbf{\circ }}$ and for radiation in

(a) the microwave range, with$\mathit{\lambda }\mathbf{=}\mathbf{3}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$ ;

(b) the visible range, with $\mathit{\lambda }\mathbf{=}\mathbf{500}\mathbf{\text{\hspace{0.17em}nm}}$;

(c) the x-ray range, with$\mathit{\lambda }\mathbf{=}\mathbf{25}\mathbf{\text{\hspace{0.17em}pm}}$ ; and

(d) the gamma-ray range, with a gamma photon energy of 1.0 MeV.

(e) What are your conclusions about the feasibility of detecting the Compton shift in these various regions of the electromagnetic spectrum, judging solely by the criterion of energy loss in a single photon-electron encounter?

(a)

The fractional change is written as follows:

$\begin{array}{c}\frac{\Delta E}{E}=\frac{\Delta \left(\frac{hc}{\lambda }\right)}{\frac{hc}{\lambda }}\\ =\lambda \Delta \left(\frac{1}{\lambda }\right)\\ =\lambda \left(\frac{1}{{\lambda }^{\text{'}}}-\frac{1}{\lambda }\right)\\ =\frac{\lambda }{{\lambda }^{\text{'}}}-1\\ =\frac{\lambda }{\lambda +\Delta \lambda }-1\\ =-\frac{1}{\frac{\lambda }{\Delta \lambda }+1}\\ =\frac{1}{\frac{\lambda }{{\lambda }_C}{(1-\cos \varphi )}^{-1}+1}\end{array}$

If

$\begin{array}{c}\lambda =3.0\text{\hspace{0.17em}cm}\\ =3.0\times {10}^{10}\text{pm}\end{array}$

and$\varphi ={90}^{\circ }$

So,

$\begin{array}{c}\frac{\Delta E}{E}=-\frac{1}{\left(\frac{3.0\times {10}^{10}}{2.43}\right){(1-\cos {90}^{\circ })}^{-1}+1}\\ =-8.1\times {10}^{-11}\\ =-8.1\times {10}^{-9}\%\end{array}$

Hence, the percentage change in photon energy during collision in microwave range is $-8.1\times {10}^{-9}\%$.

(b)

Let $\lambda =500\text{\hspace{0.17em}nm}=5.0\times {10}^5\text{pm}$and $\varphi ={90}^{\circ }$, thus it gives:

$\begin{array}{c}\frac{\Delta E}{E}=-\frac{1}{\left(\frac{5.0\times {10}^5}{2.43}\right){(1-\cos {90}^{\circ })}^{-1}+1}\\ =-4.9\times {10}^{-6}\\ =-4.9\times {10}^{-4}\%\end{array}$

Hence, the percentage change in photon energy during collision in visible range is $-4.9\times {10}^{-4}\%$.

(c)

Let$\lambda =25\text{\hspace{0.17em}pm}$and $\varphi ={90}^{\circ }$, thus it gives:

$\begin{array}{c}\frac{\Delta E}{E}=-\frac{1}{\left(\frac{25}{2.43}\right){(1-\cos {90}^{\circ })}^{-1}+1}\\ =-8.9\times {10}^{-2}\\ =-8.9\%\end{array}$

Hence, the percentage change in photon energy during collision in x-ray range is$-8.9\%$ .

(d)

Let

$\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{1240\text{\hspace{0.17em}nm.eV}}{1.0MeV}\\ =1.24\times {10}^{-3}\text{nm}\\ =1.24\text{pm}\end{array}$

and , $\varphi ={90}^{\circ }$thus it gives:

$\begin{array}{c}\frac{\Delta E}{E}=-\frac{1}{\left(\frac{1.24}{2.43}\right){(1-\cos {90}^{\circ })}^{-1}+1}\\ =-0.66\\ =-66\%\end{array}$

Hence, the percentage change in photon energy during collision in gamma ray range is$-66\%$.

(e)

From the above calculation, the shorter the wavelength the greater the fractional energy change for the photon as a result of the Compton scattering. Since$\frac{\Delta E}{E}$ is virtually zero for microwave and visible light, the Compton Effect is significant only in the x-ray to gamma ray range of the electromagnetic spectrum.

Hence, the Compton shift is zero for various of regions of electromagnetic spectrum.