Chapter 27, Problem 27P

To determine

The velocity of the scattered electron.

Answer to Problem 27P

The velocity of the scattered electron is $\underset{\_}{4.45\times {10}^6\text{m/s at }62.6°\text{south of east}}$.

Explanation of Solution

Given that the wavelength of the photon is $0.14800\text{nm}$, it is travelling in the eastward direction, the wavelength of the scattered photon is $0.14900\text{nm}$. It is obtained that the angle of scattering is $54.0°$, and the south component of electron’s momentum is $3.60\times {10}^{-24}\text{kg}\cdot \text{m/s}$.

Write the expression for the south component of electron’s momentum.

  $p_{\text{e,south}}=m_{\text{e}}{v}_{\text{s}}$                                                                                                         (I)

Here, $p_{\text{e,south}}$ is the south component of electron’s momentum, $m_{\text{e}}$ is the mass of electron, and $v_{\text{s}}$ is the south component of electron’s velocity.

Write the expression for the north component of the scattered photon’s momentum.

  ${p^{\prime }}_{\text{north}}=\frac{h}{{\lambda }^{\prime }}\sin \theta$                                                                                                    (II)

Here, ${p^{\prime }}_{\text{north}}$ is the north component of scattered momentum, $h$ is the Planck’s constant, ${\lambda }^{\prime }$ is the wavelength of scattered photon, and $\theta$ is the scattering angle.

Since the south component of electron’s momentum is equal to the north component of the scattered photon’s momentum, the right hand sides of equations (I) and (II) can be equated.

  $m_{\text{e}}{v}_{\text{s}}=\frac{h}{{\lambda }^{\prime }}\sin \theta$                                                                                                    (III)

Solve equation (III) for $v_{\text{s}}$.

  $v_{\text{s}}=\frac{h}{m_{\text{e}}{\lambda }^{\prime }}\sin \theta$                                                                                                   (IV)

Write the expression for the east component of electron’s momentum.

  $p_{\text{e,east}}=m_{\text{e}}{v}_{\text{e}}$                                                                                                        (V)

Here, $p_{\text{e,east}}$ is the east component of electron’s momentum, and $v_{\text{e}}$ is the east component of electron’s velocity.

Write the expression for the east component of electron’s momentum in terms of the momentum of the photon.

  $p_{\text{e,east}}=p-{p^{\prime }}_{\text{east}}$                                                                                                  (VI)

Here, $p$ is the momentum of the incident photon, and ${p^{\prime }}_{\text{east}}$ is the east component of momentum of scattered photon.

Modify equation (VI) by replacing the momentum of photons in terms of their wavelength and using equation (V).

  $m_{\text{e}}{v}_{\text{e}}=\frac{h}{\lambda }-\frac{h}{{\lambda }^{\prime }}\cos \theta$                                                                                            (VII)

Solve equation (VII) for $v_{\text{e}}$.

  $v_{\text{e}}=\frac{h}{m_{\text{e}}}\left(\frac{1}{\lambda }-\frac{\cos \theta }{{\lambda }^{\prime }}\right)$                                                                                         (VIII)

Write the expression for the expression for the magnitude of velocity of the scattered electron.

  $v=\sqrt{v_{\text{s}}^2+v_{\text{e}}^2}$                                                                                                        (IX)

Write the expression for the angle that the scattered electrons velocity make with the east direction.

  $\varphi ={\tan }^{-1}\left(\frac{-v_{\text{s}}}{v_{\text{e}}}\right)$                                                                                                   (X)

Here, $\varphi$ is the angle that the scattered electrons velocity make with the east direction.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.109\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$, $0.14900\text{nm}$ for ${\lambda }^{\prime }$, and $54.0°$ for $\theta$ in equation (IV) to find $v_{\text{s}}$.

  $\begin{array}{c}{v}_{\text{s}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.109\times {10}^{-31}\text{kg}\right)\left(0.14900\text{nm}\right)}\sin 54.0°\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.109\times {10}^{-31}\text{kg}\right)\left(0.14900\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}\right)}\sin 54.0°\\ =3.95\times {10}^6\text{m/s}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.109\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$, $0.14800\text{nm}$ for $\lambda$, $0.14900\text{nm}$ for ${\lambda }^{\prime }$, and $54.0°$ for $\theta$ in equation (VIII) to find $v_{\text{e}}$.

  $\begin{array}{c}{v}_{\text{e}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{9.109\times {10}^{-31}\text{kg}}\left(\frac{1}{0.14800\text{nm}}-\frac{\cos 54.0°}{0.14900\text{nm}}\right)\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{9.109\times {10}^{-31}\text{kg}}\left(\frac{1}{0.14800\text{nm}}-\frac{\cos 54.0°}{0.14900\text{nm}}\right)\left(\frac{1\times {10}^9\text{nm}}{1\text{m}}\right)\\ =2.05\times {10}^6\text{m/s}\end{array}$

Substitute $3.95\times {10}^6\text{m/s}$ for $v_{\text{s}}$, and $2.05\times {10}^6\text{m/s}$ for $v_{\text{e}}$ in equation (IX) to find $v$.

  $\begin{array}{c}v=\sqrt{{\left(3.95\times {10}^6\text{m/s}\right)}^2+{\left(2.05\times {10}^6\text{m/s}\right)}^2}\\ =4.45\times {10}^6\text{m/s}\end{array}$

Substitute $3.95\times {10}^6\text{m/s}$ for $v_{\text{s}}$, and $2.05\times {10}^6\text{m/s}$ for $v_{\text{e}}$ in equation (X) to find $\varphi$.

  $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{-3.95\times {10}^6\text{m/s}}{2.05\times {10}^6\text{m/s}}\right)\\ =-62.6°\end{array}$

This indicates that the electron travels in a direction $62.6°$ south of east.

Therefore, the velocity of the scattered electron is $\underset{\_}{4.45\times {10}^6\text{m/s at }62.6°\text{south of east}}$.