Chapter 40, Problem 47P

(a)

To determine

The ratio of the wavelength of the photon to the wavelength of the electron.

Answer to Problem 47P

The ratio of the wavelength of the photon to the wavelength of the electron is $\frac{\gamma }{\gamma -1}\frac{u}{c}$ .

Explanation of Solution

It is given that the speed of the electron is close to the speed of light. This implies relativistic expression for energy has to be used.

Write the equation for the kinetic energy of the electron.

  $K=\left(\gamma -1\right)m_e{c}^2$                                                                                                          (I)

Here, $K$ is the kinetic energy of the electron, $\gamma$ is the Lorentz factor, $m_e$ is the mass of the electron and $c$ is the speed of light in vacuum.

Write the equation for the de Broglie wavelength of the electron.

  $\lambda =\frac{h}{p}$                                                                                                                      (II)

Here, $\lambda$ is the de Broglie wavelength of the electron, $h$ is the Planck’s constant and $p$ is the momentum of the electron.

Write the equation for the momentum of the electron.

  $p=\gamma m_{e}u$

Here, $m_e$ is the mass of the electron and $u$ is the speed of the electron.

  $\lambda =\frac{h}{\gamma m_{e}u}$                                                                                                               (III)

It is given that energy of the photon is equal to the kinetic energy of the electron.

Write the relationship between the energy of the photon and the kinetic energy of the electron.

  $E=K$                                                                                                                    (IV)

Here, $E$ is the energy of the photon.

Write the equation for the energy of the photon in terms of its frequency.

  $E=hf$                                                                                                                      (V)

Here, $f$ is the frequency of the photon.

Write the equation for the wavelength of the photon.

  ${\lambda }_{\text{ph}}=\frac{c}{f}$

Here, ${\lambda }_{\text{ph}}$ is the wavelength of the photon.

Multiply the numerator and denominator of the right hand side of the above equation by $h$ .

  ${\lambda }_{\text{ph}}=\frac{ch}{hf}$

Replace denominator of the right hand side of the above equation by equation (V).

  ${\lambda }_{\text{ph}}=\frac{ch}{E}$

Put equation (IV) in the above equation.

  ${\lambda }_{\text{ph}}=\frac{ch}{K}$

Put equation (I) in the above equation.

  ${\lambda }_{\text{ph}}=\frac{ch}{\left(\gamma -1\right)m_e{c}^2}$                                                                                                    (VI)

Conclusion:

Take the ratio of equation (VI) to equation (III).

  $\begin{array}{c}\frac{{\lambda }_{\text{ph}}}{\lambda }=\frac{\frac{ch}{\left(\gamma -1\right)m_e{c}^2}}{\frac{h}{\gamma m_{e}u}}\\ =\frac{ch}{\left(\gamma -1\right)m_e{c}^2}\frac{\gamma m_{e}u}{h}\\ =\frac{\gamma }{\gamma -1}\frac{u}{c}\end{array}$                                                                                        (VII)

Therefore, the ratio of the wavelength of the photon to the wavelength of the electron is $\frac{\gamma }{\gamma -1}\frac{u}{c}$ .

(b)

To determine

The value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.900c$ .

Answer to Problem 47P

The value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.900c$ is $1.60$ .

Explanation of Solution

Write the expression for the Lorentz factor.

  $\gamma =\frac{1}{\sqrt{1-u^2/c^2}}$

Put the above equation in equation (VII).

  $\frac{{\lambda }_{\text{ph}}}{\lambda }=\frac{1}{\sqrt{1-u^2/c^2}}\frac{1}{\left[\left(1/\sqrt{1-u^2/c^2}\right)-1\right]}\frac{u}{c}$                                                             (VIII)

Conclusion:

Substitute $0.900c$ for $u$ in equation (VIII) to find the value of the ratio.

  $\begin{array}{c}\frac{{\lambda }_{\text{ph}}}{\lambda }=\frac{1}{\sqrt{1-{\left(0.900c\right)}^2/c^2}}\frac{1}{\left(1/\sqrt{1-{\left(0.900c\right)}^2/c^2}\right)-1}\frac{0.900c}{c}\\ =\frac{1}{\sqrt{1-{\left(0.900\right)}^2}}\frac{0.900}{\left(1/\sqrt{1-{\left(0.900\right)}^2}\right)-1}\\ =1.60\end{array}$

Therefore, the value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.900c$ is $1.60$ .

(c)

To determine

The change in answer of part (b) if the particle were a proton instead of electron.

Answer to Problem 47P

There will be no change in the answer to part (b) since the ration does not depend on mass.

Explanation of Solution

The expression for the ration of the wavelength of the photon to the wavelength of the material particle is given by equation (VIII). The equation contains only the speed of the material particle and the speed of light in vacuum. The equation is independent of the mass of the particle.

Since the ratio is independent of the mass of the material particle, the value of the ratio will be the same if the electron is replaced with a proton of same speed. This implies there will be no change in the answer to part (b) even when the electron is replaced with a proton.

Conclusion:

Thus, there will be no change in the answer to part (b) since the ration does not depend on mass.

(d)

To determine

The value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.00100c$ .

Answer to Problem 47P

The value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.00100c$ is $2.00\times {10}^3$ .

Explanation of Solution

Equation (VIII) can be used to find the value of the ratio.

Conclusion:

Substitute $0.00100c$ for $u$ in equation (VIII) to find the value of the ratio.

  $\begin{array}{c}\frac{{\lambda }_{\text{ph}}}{\lambda }=\frac{1}{\sqrt{1-{\left(0.00100c\right)}^2/c^2}}\frac{1}{\left(1/\sqrt{1-{\left(0.00100c\right)}^2/c^2}\right)-1}\frac{0.00100c}{c}\\ =\frac{1}{\sqrt{1-{\left(0.00100\right)}^2}}\frac{0.00100}{\left(1/\sqrt{1-{\left(0.00100\right)}^2}\right)-1}\\ =2.00\times {10}^3\end{array}$

Therefore, the value of the ratio of the wavelength of the photon to the wavelength of the electron for the particle speed $u=0.00100c$ is $2.00\times {10}^3$ .

(e)

To determine

The value the ratio of the wavelengths approaches at high particle speeds.

Answer to Problem 47P

The value the ratio of the wavelengths approaches at high particle speeds is $1$ .

Explanation of Solution

At high particles speeds, $u\to c$ . As $u\to c$, $\frac{u}{c}\to 1$ .

Substitute this in the expression for Lorentz factor.

  $\begin{array}{c}\gamma \to \frac{1}{\sqrt{1-1}}\\ \gamma \to \infty \end{array}$

As $\gamma \to \infty ,\gamma -1\to \gamma$ .

Conclusion:

Replace $\gamma -1$ by $\gamma$ and $\frac{u}{c}$ by $1$ in equation (VII) to find the value of the ratio at high particle speeds.

  $\begin{array}{c}\frac{{\lambda }_{\text{ph}}}{\lambda }\to \frac{\gamma }{\gamma }\left(1\right)\\ =1\end{array}$

Therefore, the value the ratio of the wavelengths approaches at high particle speeds is $1$ .

(f)

To determine

The value the ratio of the wavelengths approaches at low particle speeds.

Answer to Problem 47P

The value the ratio of the wavelengths approaches at low particle speeds is $\infty$ .

Explanation of Solution

At low particles speeds, $\frac{u}{c}\to 0$ .

As $\frac{u}{c}\to 0$,

  $\begin{array}{c}\gamma ={\left(1-\frac{u^2}{c^2}\right)}^{-1/2}\to {\left(1-0\right)}^{-1/2}\\ \gamma \to 1\end{array}$                                                                                  (IX)

Also,

  $\begin{array}{c}\gamma -1={\left(1-\frac{u^2}{c^2}\right)}^{-1/2}-1\to 1-\left(-\frac{1}{2}\right)\frac{u^2}{c^2}-1\\ =\frac{1}{2}\frac{u^2}{c^2}\end{array}$                                                                 (X)

Conclusion:

Put approximations (IX) and (X) in equation (VII) to find the value of the ratio at low particle speeds.

  $\begin{array}{c}\frac{{\lambda }_{\text{ph}}}{\lambda }\to 1\frac{u/c}{\left(1/2\right)\left(u^2/c^2\right)}\\ =\frac{2c}{u}\end{array}$

For low values of $u$ , $\frac{2c}{u}\to \infty$ .

Replace $\frac{2c}{u}$ by this approximation in the above equation.

  $\frac{{\lambda }_{\text{ph}}}{\lambda }\to \infty$

Therefore, the value the ratio of the wavelengths approaches at low particle speeds is $\infty$ .