Chapter 27, Problem 28P

(a)

To determine

The wavelength of the incident photon.

Answer to Problem 28P

The wavelength of the incident photon is $0.101\text{nm}$.

Explanation of Solution

The Compton shift formula is,

$\lambda \text{'}-\lambda =\frac{h}{m_{e}c}\left(1-\cos \theta \right)$ (1)

• • $h$ is the Plank’s constant
  • $c$ is the speed of the light
  • $m_e$ is the mass of the electron

The kinetic energy of the recoiling electron, according to the conservation of energy is,

$\begin{array}{c}\frac{1}{2}{m}_e{v}^2=\frac{hc}{{\lambda }_0}-\frac{hc}{\lambda }\\ =\frac{hc\left(\lambda -{\lambda }_0\right)}{{\lambda }_0\lambda }\end{array}$

Using equation (1) in this equation,

$\begin{array}{c}\frac{1}{2}{m}_e{v}^2=\frac{hc\left[\frac{h}{m_{e}c}\left(1-\cos \theta \right)\right]}{{\lambda }_0\left[{\lambda }_0+\frac{h}{m_{e}c}\left(1-\cos \theta \right)\right]}\\ {\lambda }_0+\frac{h}{m_{e}c}\left(1-\cos \theta \right)=2{\left(\frac{h}{m_{e}c}\right)}^2\left(1-\cos \theta \right)\end{array}$

On simplification, the equation is seen to be in the form $a{\lambda }_0^2+b{\lambda }_0+c=0$, where $a=1$.

$b=\frac{h}{m_{e}c}\left(1-\cos \theta \right)$

Substitute $17.4°$ for $\theta$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ and $3.00\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}b=\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(3.00\times {10}^8\text{m/s}\right)}\left(1-\cos 17.4°\right)\\ =1.11\times {10}^{-13}\text{m}\end{array}$

$c=-2{\left(\frac{h}{m_{e}v}\right)}^2\left(1-\cos \theta \right)$

Substitute $17.4°$ for $\theta$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ and $2.18\times {10}^6\text{m/s}$ for $v$.

$\begin{array}{c}c=\frac{-2{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)}^2}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.18\times {10}^6\text{m/s}\right)}\left(1-\cos 17.4°\right)\\ =-1.02\times {10}^{-20}{\text{m}}^2\end{array}$

Solving the quadratic equation, the wavelength of the incident photon is, ${\lambda }_0=0.101\text{nm}$

Conclusion:

Thus, the wavelength of the incident photon is $0.101\text{nm}$.

(b)

To determine

The angle through which the electron scatters.

Answer to Problem 28P

The angle through which the electron scatters is $80.9°$.

Explanation of Solution

Conserving momentum in the y-axis,

$p_e\sin \varphi =p\sin \theta$

The wavelength of the scattered photon is,

$\lambda ={\lambda }_0+\frac{h}{m_{e}c}\left(1-\cos \theta \right)$

Substitute $17.4°$ for $\theta$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $1.01\times {10}^{-10}\text{m}$ for $\lambda$ and $3.00\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}\lambda =1.01\times {10}^{-10}\text{m}+\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(1-\cos 17.4°\right)}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(3.00\times {10}^8\text{m/s}\right)}\\ =1.011\times {10}^{-10}\text{m}\end{array}$

The scattering angle for the electron is,

$\varphi ={\sin }^{-1}\left[\frac{h\sin \theta }{\lambda m_{e}v}\right]$

Substitute $17.4°$ for $\theta$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $1.01\times {10}^{-10}\text{m}$ for $\lambda$ and $2.18\times {10}^6\text{m/s}$ for $v$.

$\begin{array}{c}\varphi ={\sin }^{-1}\left[\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\sin 17.0°}{\left(1.011\times {10}^{-10}\text{m}\right)\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.18\times {10}^6\text{m/s}\right)}\right]\\ =80.9°\end{array}$

Conclusion:

Thus, the angle through which the electron scatters is $80.9°$