Chapter 3, Problem 50P

(a)

To determine

The wavelength of the incident photon to scatter at the angle of $30°$.

Answer to Problem 50P

The wavelength of the incident photon that scatters the photons at angle of $30°$ is $0.0813\text{nm}$.

Explanation of Solution

Write the expression for the Crompton scattering of the light.

  $\Delta \lambda =\frac{h}{mc}\left(1-\cos \theta \right)$        (I)

Here, $\Delta \lambda$ is the change in the wavelength, $h$ is the Planck’s constant, $m$ is the mass of the proton, $\theta$ is the angle of scattering of electron and $c$ is the speed of light.

Write the expression for the wavelength of the photon required.

  $\lambda =\frac{\Delta \lambda }{4\times {10}^{-3}}$        (II)

Here, $\lambda$ is the wavelength of the photons.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.10\times {10}^{-31}\text{kg}$ for $m$, $30°$ for $\theta$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I).

  $\begin{array}{c}\Delta \lambda =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.10\times {10}^{-31}\text{kg}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\left(1-\cos 30°\right)\\ =\frac{6.626\times {10}^{-34}}{2.73\times {10}^{-22}}\left(0.134\right)\\ =3.25\times {10}^{-13}\text{m}\end{array}$

Substitute $3.25\times {10}^{-13}\text{m}$ for $\Delta \lambda$ in equation (II).

  $\begin{array}{c}\lambda =\frac{3.25\times {10}^{-13}\text{m}}{4\times {10}^{-3}}\\ =8.13\times {10}^{-11}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =0.0813\text{nm}\end{array}$

Thus, the wavelength of the incident photon that scatters the photons at angle of $30°$ is $0.0813\text{nm}$.

(b)

To determine

The wavelength of the incident photon to scatter at the angle of $90°$.

Answer to Problem 50P

The wavelength of the incident photon that scatters the photons at angle of $90°$ is $0.608\text{nm}$.

Explanation of Solution

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.10\times {10}^{-31}\text{kg}$ for $m$, $90°$ for $\theta$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I).

  $\begin{array}{c}\Delta \lambda =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.10\times {10}^{-31}\text{kg}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\left(1-\cos 90°\right)\\ =\frac{6.626\times {10}^{-34}}{2.73\times {10}^{-22}}\\ =2.432\times {10}^{-12}\text{m}\end{array}$

Substitute $2.432\times {10}^{-12}\text{m}$ for $\Delta \lambda$ in equation (II).

  $\begin{array}{c}\lambda =\frac{2.432\times {10}^{-12}\text{m}}{4\times {10}^{-3}}\\ =6.08\times {10}^{-10}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =0.608\text{nm}\end{array}$

Thus, the wavelength of the incident photon that scatters the photons at angle of $90°$ is $0.608\text{nm}$.

(c)

To determine

The wavelength of the incident photon to scatter at the angle of $170°$.

Answer to Problem 50P

The wavelength of the incident photon that scatters the photons at angle of $170°$ is $1.206\text{nm}$.

Explanation of Solution

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.10\times {10}^{-31}\text{kg}$ for $m$, $170°$ for $\theta$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I).

  $\begin{array}{c}\Delta \lambda =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.10\times {10}^{-31}\text{kg}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\left(1-\cos 170°\right)\\ =\frac{6.626\times {10}^{-34}}{2.73\times {10}^{-22}}\left(1.985\right)\\ =4.827\times {10}^{-12}\text{m}\end{array}$

Substitute $4.827\times {10}^{-12}\text{m}$ for $\Delta \lambda$ in equation (II).

  $\begin{array}{c}\lambda =\frac{4.827\times {10}^{-12}\text{m}}{4\times {10}^{-3}}\\ =1.206\times {10}^{-10}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1.206\text{nm}\end{array}$

Thus, the wavelength of the incident photon that scatters the photons at angle of $170°$ is $1.206\text{nm}$.