91Ó°ÊÓ

(a)

The fractional change is written as follows:

$\begin{array}{c}\frac{\mathrm{Δ}E}{E}=\frac{\mathrm{Δ}\left(\frac{hc}{\mathrm{λ}}\right)}{\frac{hc}{\mathrm{λ}}}\\ =\mathrm{λ}\mathrm{Δ}\left(\frac{1}{\mathrm{λ}}\right)\\ =\mathrm{λ}\left(\frac{1}{{\mathrm{λ}}^{\text{'}}}−\frac{1}{\mathrm{λ}}\right)\\ =\frac{\mathrm{λ}}{{\mathrm{λ}}^{\text{'}}}−1\\ =\frac{\mathrm{λ}}{\mathrm{λ}+\mathrm{Δ}\mathrm{λ}}−1\\ =−\frac{1}{\frac{\mathrm{λ}}{\mathrm{Δ}\mathrm{λ}}+1}\\ =\frac{1}{\frac{\mathrm{λ}}{{\mathrm{λ}}_C}{(1−\cos \mathrm{ϕ})}^{−1}+1}\end{array}$

If

$\begin{array}{c}\mathrm{λ}=3.0\text{ c³¾}\\ =3.0×{10}^{10}\text{pm}\end{array}$

and$\mathrm{ϕ}={90}^{∘}$

So,

$\begin{array}{c}\frac{\mathrm{Δ}E}{E}=−\frac{1}{\left(\frac{3.0×{10}^{10}}{2.43}\right){(1−\cos {90}^{∘})}^{−1}+1}\\ =−8.1×{10}^{−11}\\ =−8.1×{10}^{−9}\%\end{array}$

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

(b)

Let $\mathrm{λ}=500\text{ n³¾}=5.0×{10}^5\text{pm}$and $\mathrm{Ï•}={90}^{∘}$, thus it gives:

$\begin{array}{c}\frac{\mathrm{Δ}E}{E}=−\frac{1}{\left(\frac{5.0×{10}^5}{2.43}\right){(1−\cos {90}^{∘})}^{−1}+1}\\ =−4.9×{10}^{−6}\\ =−4.9×{10}^{−4}\%\end{array}$

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

(c)

Let$\mathrm{λ}=25\text{ p³¾}$and $\mathrm{Ï•}={90}^{∘}$, thus it gives:

$\begin{array}{c}\frac{\mathrm{Δ}E}{E}=−\frac{1}{\left(\frac{25}{2.43}\right){(1−\cos {90}^{∘})}^{−1}+1}\\ =−8.9×{10}^{−2}\\ =−8.9\%\end{array}$

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

(d)

Let

$\begin{array}{c}\mathrm{λ}=\frac{hc}{E}\\ =\frac{1240\text{ n³¾.eV}}{1.0MeV}\\ =1.24×{10}^{−3}\text{nm}\\ =1.24\text{pm}\end{array}$

and , $\mathrm{ϕ}={90}^{∘}$thus it gives:

$\begin{array}{c}\frac{\mathrm{Δ}E}{E}=−\frac{1}{\left(\frac{1.24}{2.43}\right){(1−\cos {90}^{∘})}^{−1}+1}\\ =−0.66\\ =−66\%\end{array}$

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$\frac{\mathrm{Δ}E}{E}$ 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.