Chapter 27, Problem 32P

a.

To determine

Compton wavelength shift for protons detected at $45°$ angle.

Answer to Problem 32P

Solution:

Compton wavelength shift for protons detected at $45°$ angle is $3.87\times {10}^{-14}\text{m}$ .

Explanation of Solution

Given:

Mass of the proton    : $1.67\times {10}^{-27}\text{kg}$

Speed of the light      : $3.0\times {10}^8\text{m/s}$

Value of plank’s constant  : $6.626\times {10}^{-34}{\text{m}}^{\text{2}}\text{Kg/s}$

Formula Used:

${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$

Calculation:

By using the formula, ${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$ thecompton wavelength shift for protons can be calculated as

$\begin{array}{c}{\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)\end{array}$

   $\begin{array}{c}=\frac{6.626\times {10}^{-34}\text{J}\text{s}}{1.67\times {10}^{-27}\text{kg}\times 3.0\times {10}^8\text{m/s}}\left(1-\cos 45°\right)\end{array}$

   $\begin{array}{c}=1.32\times {10}^{-15}\text{m}\left(1-\cos 45°\right)\end{array}$

   $\begin{array}{c}=3.87\times {10}^{-14}\text{m}\end{array}$

Conclusion:

Compton wavelength shift for protons detected at $45°$ angle is $3.87\times {10}^{-14}\text{m}$ .

b.

To determine

Compton wavelength shift for protons detected at $90°$ angle.

Answer to Problem 32P

Solution:

Compton wavelength shift for protons detected at $90°$ angle is $1.32\times {10}^{-15}\text{m}$ .

Explanation of Solution

Given:

Mass of the proton    : $1.67\times {10}^{-27}\text{kg}$

Speed of the light    : $3.0\times {10}^8\text{m/s}$

Value of plank’s constant  : $6.626\times {10}^{-34}{\text{m}}^{\text{2}}\text{Kg/s}$

Formula Used:

${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$

Calculation:

By using the formula, ${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$ the compton wavelength shift for protons can be calculated as

$\begin{array}{c}{\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)\end{array}$

   $\begin{array}{c}=\frac{6.626\times {10}^{-34}\text{J}\text{s}}{1.67\times {10}^{-27}\text{kg}\times 3.0\times {10}^8\text{m/s}}\left(1-\cos 90°\right)\end{array}$

   $\begin{array}{c}=1.32\times {10}^{-15}\text{m}\left(1-\cos 90°\right)\end{array}$

   $\begin{array}{c}=1.32\times {10}^{-15}\text{m}\end{array}$

Conclusion:

Compton wavelength shift for protons detected at $90°$ angle is $1.32\times {10}^{-15}\text{m}$ .

c.

To determine

Compton wavelength shift for protons detected at $180°$ angle.

Answer to Problem 32P

Solution:

Compton wavelength shift for protons detected at $180°$ angle is $2.64\times {10}^{-15}\text{m}$

Explanation of Solution

Given:

Mass of the proton    : $1.67\times {10}^{-27}\text{kg}$

Speed of the light    : $3.0\times {10}^8\text{m/s}$

Value of plank’s constant  : $6.626\times {10}^{-34}{\text{m}}^{\text{2}}\text{Kg/s}$

Formula Used:

${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$

Calculation:

By using the formula, ${\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)$ the compton wavelength shift for protons can be calculated as

$\begin{array}{c}{\lambda }^{\prime }-\lambda =\left(\frac{h}{m_{0}c}\right)\left(1-\cos \theta \right)\end{array}$

   $\begin{array}{c}=\frac{6.626\times {10}^{-34}\text{J}\text{s}}{1.67\times {10}^{-27}\text{kg}\times 3.0\times {10}^8\text{m/s}}\left(1-\cos 180°\right)\end{array}$

   $\begin{array}{c}=1.32\times {10}^{-15}\text{m}\left(1-\cos 180°\right)\end{array}$

   $\begin{array}{c}=2.64\times {10}^{-15}\text{m}\end{array}$

Conclusion:

Compton wavelength shift for protons detected at $180°$ angle is $2.64\times {10}^{-15}\text{m}$ .