A photon scatters off of a free electron. (a) What is the maximum possible change in wavelength? (b) Suppose a photon scatters off of a free proton. What is the maximum possible change in wavelength now? (e) Which more clearly demonstrates the particle nature of electromagnetic radiation--collision with an electron or collision with a proton?

The change in the wavelength is directly proportional to the Planck’s constant and the angle subtended by a particle. Moreover, the wavelength is inversely proportional to the light’s speed and the mass of an electron.

During the scattering, the maximum wavelength change occurs at an angle of .

The equation of the maximum possible change in the wavelength of the electron is expressed as:

${\lambda }^{\text{'}}-\lambda =\frac{h}{m_{e}C}(1-\cos \theta )$

Here, role="math" localid="1657610341456" ${\lambda }^{\text{'}}-\lambda$is the maximum change in the wavelength, $h$is the Planck’s constant, $m_e$is the mass of an electron and $\theta$is the angle subtended by the proton.

Substitute $6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\text{for}h,9.1\times {10}^{-31}\mathrm{kg}\text{for}{m}_{\mathrm{e}},3\times {10}^8\mathrm{m}/\mathrm{s}\text{for}c\text{and}180°$

for $\theta$in the above equation.

role="math" localid="1657610242005" $$\begin{gathered} {\lambda }^{\text{'}}-\lambda =\frac{6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(9.1\times {10}^{-31}\mathrm{kg}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}\left(1-\cos 180°\right) \\ =\frac{6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(2.73\times {10}^{-22}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)}(1-(-1\left)\right) \\ =\frac{1.326\times {10}^{-33}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(2.73\times {10}^{-22}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =4.85\times {10}^{-3}\mathrm{m} \end{gathered}$$

Thus, the maximum possible change in wavelength of the electron is$=4.85\times {10}^{-3}\mathrm{m}$ .

During the scattering, the maximum wavelength change occurs at an angle of$180°$.

The equation of the maximum possible change in the wavelength of the proton is expressed as:

${\lambda }^{\text{'}}-\lambda =\frac{h}{m_{p}c}(1-\cos \theta )$

Here, ${\lambda }^{\text{'}}-\lambda$is the maximum change in the wavelength, $h$is the Planck’s constant, $m_p$is the mass of an electron and $\theta$is the angle subtended by the proton.

Substitute $6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\text{for}h,1.67\times {10}^{-27}\mathrm{kg}\text{for}{m}_e,3\times {10}^8\mathrm{m}/\mathrm{s}\text{for}c$and ${180}^{°}$

in the above equation.

$$\begin{gathered} {\lambda }^{\text{'}}-\lambda =\frac{6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}\left(1-\cos 180°\right) \\ =\frac{6.63\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(5.01\times {10}^{-19}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)}(1-(-1\left)\right) \\ =\frac{1.326\times {10}^{-33}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(5.01\times {10}^{-19}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =2.646\times {10}^{-15}\mathrm{m} \end{gathered}$$

Thus, the maximum possible change in wavelength of the proton isrole="math" localid="1657610571650" $2.646\times {10}^{-15}\mathrm{m}$

The collision with a particular electron more clearly demonstrates the particle nature of electromagnetic radiation as the wavelength change in noticeable. In the wavelength of the proton, there is an insignificant wavelength change. The photon in the x-ray band mainly interacts with other proton as a wave function and also the scattered protons also have a same type of wavelength.

Thus, the collision with an electron more clearly demonstrates the particle nature of electromagnetic radiation.