An electron moving to the left at $\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{c}$ collides with an incoming photon moving to the right . After the collision, the electron is moving to the right at 0.6c and an outgoing photon moves to the left. What was the wavelength of the incoming photon?

The relativistic energy E for a particle with mass m is

$\mathbf{E}\mathbf{=}{\mathbf{\gamma }}_n{\mathbf{mc}}^2$

Here, c is the speed of light in vacuum, and${\gamma }_n$is the Lorentz factor.

${\gamma }_n=\frac{1}{\sqrt{1-{\left(\frac{\mathrm{u}}{\mathrm{c}}\right)}^2}}$

Here, velocity of the particle u.

The relativistic momentum p of an object of mass m and velocity u is

$\rho ={\gamma }_{n}mu$

Here, ${\mathrm{\gamma }}_{\mathrm{n}}$being Lorentz factor

The energy E of a photon of wavelength$\mathrm{\lambda }$is:

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

Here, h is the Plank’s Constant $6.6\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$. And c is the speed of light

If the electron is moving, it will start the interaction with some momentum and energy already.

Momentum of the electron and photon in the initial and final stage is:

${\mathrm{P}}_{\mathrm{pi}}+{\mathrm{P}}_{\mathrm{ei}}={\mathrm{P}}_{\mathrm{pf}}+{\mathrm{P}}_{\mathrm{ef}}$ ….. (1)

Here P refers to momentum, the suffix e and p refer to proton and electron.

The momentum of the photon in the initial state:

${\mathrm{P}}_{\mathrm{pi}}=\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}$

Similarly the momentum of the photon in the final state,

${\mathrm{P}}_{\mathrm{ei}}=-\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}$

The momentum of the electron in the initial state is,

$P_{ei}={\gamma }_{i}m{u}_i$

The momentum of the electron in the final state is,

${\mathrm{P}}_{\mathrm{ef}}={\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}$

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photons momentum after the collision:

Rearranging the equation (1), we get

${\mathrm{P}}_{\mathrm{pi}}+{\mathrm{P}}_{\mathrm{ei}}=-{\mathrm{P}}_{\mathrm{pf}}+{\mathrm{P}}_{\mathrm{ef}}$

Substitute $\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}\mathrm{for}{\mathrm{P}}_{\mathrm{pi}},-\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}\mathrm{for}{\mathrm{P}}_{\mathrm{pf}},{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}\mathrm{for}{\mathrm{P}}_{\mathrm{pi}},{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\mathrm{for}{\mathrm{P}}_{\mathrm{ef}}$ in the equation (1) and solve,

$\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}=-\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}$ ….. (2)

Next write out the energy conservation equation and expand it:

${\mathrm{E}}_{\mathrm{pi}}+{\mathrm{E}}_{\mathrm{ei}}={\mathrm{E}}_{\mathrm{pf}}+{\mathrm{E}}_{\mathrm{ei}}$

Kinetic energy of the electron and photon in the initial stage is:

${\mathrm{E}}_{\mathrm{p}}+{\mathrm{E}}_{\mathrm{ei}}={\mathrm{E}}_{\mathrm{ef}}$ ….. (3)

The energy of the electron in the initial stage is,

${\mathrm{E}}_{\mathrm{pi}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{{}_{\mathrm{i}}}}$

The energy of the electron in the final stage is,

${\mathrm{E}}_{\mathrm{pi}}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{f}}}$

Energy of the photon in the initial stage is,

${\mathrm{E}}_{\mathrm{ei}}={\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2$

Here${\mathrm{\gamma }}_{\mathrm{i}}$is the frequency of the photon in the initial stage Energy of the electron in the final stage is,${\mathrm{E}}_{\mathrm{ef}}={\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mc}}^2$

Here ${\mathrm{\gamma }}_{\mathrm{f}}$ is the frequency of the photon in the final state,

Substitute$\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}\mathrm{fo}{\mathrm{rE}}_{\mathrm{pi}},-\frac{\mathrm{h}}{{\mathrm{\lambda }}_1}\mathrm{for}{\mathrm{E}}_{\mathrm{pf}},{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}\mathrm{for}{\mathrm{E}}_{\mathrm{pi}},{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\mathrm{for}{\mathrm{E}}_{\mathrm{ef}}$in the equation (3)

$\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2=-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2$ …… (4)

Solve the equation for $\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{f}}},$

$\begin{array}{c}\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2=-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2\\ \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}+{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}}}+{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mc}}^2\\ \frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{f}}}=\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+({\mathrm{\gamma }}_{\mathrm{i}}\mathrm{c}-{\mathrm{\gamma }}_{\mathrm{f}}\mathrm{c})\mathrm{m}\end{array}$

Substitute$\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+({\mathrm{\gamma }}_{\mathrm{i}}\mathrm{c}-{\mathrm{\gamma }}_{\mathrm{f}}\mathrm{c})\mathrm{m}$for$\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{f}}},$in the equation (2) and solve,

$\begin{array}{l}\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}=-\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\\ \frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}=-\left[\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+({\mathrm{\gamma }}_{\mathrm{i}}-{\mathrm{\gamma }}_{\mathrm{f}})\mathrm{mc}\right]+{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\\ \frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}=-\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+({\mathrm{\gamma }}_{\mathrm{f}}-{\mathrm{\gamma }}_{\mathrm{i}})\mathrm{mc}+{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\end{array}$

Then get all the ${\mathrm{\gamma }}_{\mathrm{i}}$ terms by themselves on one side and combine like terms:

$\begin{array}{c}\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}-{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}=-\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+({\mathrm{\gamma }}_{\mathrm{f}}-{\mathrm{\gamma }}_{\mathrm{i}})\mathrm{mc}+{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}\\ \frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}+\frac{\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{f}}}={\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}+({\mathrm{\gamma }}_{\mathrm{i}}-{\mathrm{\gamma }}_{\mathrm{f}})\mathrm{mc}\\ \frac{2\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}=\mathrm{m}[{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}+{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{u}}_{\mathrm{f}}+({\mathrm{\gamma }}_{\mathrm{i}}-{\mathrm{\gamma }}_{\mathrm{f}})\mathrm{c}]\end{array}$

And just rearrange terms so as to solve for the initial wavelength ${\mathrm{\lambda }}_{\mathrm{i}}$.

$\frac{2\mathrm{h}}{{\mathrm{\lambda }}_{\mathrm{i}}}=\mathrm{m}[{\mathrm{\gamma }}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}+{\mathrm{\gamma }}_{\mathrm{f}}{\mathrm{u}}_{\mathrm{f}}+({\mathrm{\gamma }}_{\mathrm{i}}-{\mathrm{\gamma }}_{\mathrm{f}})\mathrm{c}]$

The c was factored out so as to make calculations later a litter easier. It would be helpful to calculate the Lorentz factors so that they can easily plugged into the formula:

${\mathrm{\gamma }}_{\mathrm{i}}=\frac{1}{\sqrt{1-{\left(\frac{{\mathrm{u}}_{\mathrm{i}}}{\mathrm{c}}\right)}^2}}$

Substituting 0.8c for $u_i$and solve,

$\begin{array}{c}{\gamma }_i=\frac{1}{\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}}\\ =\frac{5}{3}\end{array}$

Substituting 0.6c for $u_i$and solve,

$\begin{array}{c}{\gamma }_f=\frac{1}{\sqrt{1-{\left(\frac{0.6c}{c}\right)}^2}}\\ =1.25\end{array}$

Substitute$6.63\times {10}^{-34}\text{\hspace{0.17em}J-s}$in for h, $9\times {10}^{-31}\text{\hspace{0.17em}m}$in for m,5/3 in for ${\gamma }_i$, $0.8c$ for u, 1.25 in for ${\gamma }_i$, $\text{0.6c}$for u and solve asL

${\mathrm{\lambda }}_{\mathrm{i}}=\frac{2\mathrm{h}}{\mathrm{mc}\left[{\mathrm{\gamma }}_{\mathrm{i}}\frac{{\mathrm{u}}_{\mathrm{i}}}{\mathrm{c}}+{\mathrm{\gamma }}_{\mathrm{f}}\frac{{\mathrm{u}}_{\mathrm{f}}}{\mathrm{c}}+({\mathrm{\gamma }}_{\mathrm{f}}-{\mathrm{\gamma }}_{\mathrm{i}})\right]}$

So the initial wavelength of the photon of ${\text{Pb}}^{\text{2+}}$ion is $2.9\times {10}^{-12}\text{\hspace{0.17em}m}$.