Chapter 27, Problem 31P

To determine

(a) To Determine:

The wavelength of the proton.

Answer to Problem 31P

Solution:

The wavelength of the proton is $2.43\times {10}^{-12}\text{m}$.

Explanation of Solution

Given:

Mass of the proton: $9.11\times {10}^{-31}\text{kg}$

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:

$\lambda =\frac{h}{mc}$

Calculation:

By using the formula, $\lambda =\frac{h}{mc}$

We can find the Compton wavelength for proton

$\begin{array}{l}{\lambda }_{compton}=\frac{h}{mc}\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{9.11\times {10}^{-31}\text{kg}\times 3.0\times {10}^8\text{m/s}}=2.43\times {10}^{-12}\text{m}\\ \end{array}$

To determine

(b) To Determine:

The wavelength of the proton.

Answer to Problem 31P

Solution:

The wavelength of the proton 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:

$\lambda =\frac{h}{mc}$

Calculation:

By using the formula, $\lambda =\frac{h}{mc}$

We can find the Compton wavelength for proton

$\begin{array}{l}{\lambda }_{compton}=\frac{h}{mc}\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{1.67\times {10}^{-27}\text{kg}\times 3.0\times {10}^8\text{m/s}}=1.32\times {10}^{-15}\text{m}\\ \end{array}$

To determine

(c) To Determine:

Validate that the photon’s energy is equal to the rest energy of the particle.

Answer to Problem 31P

Solution:

If photon’s wavelength is equal to the Compton wavelength of a particle, then the photon’s energy is equal to the rest energy of the particle that is $E=m_0{c}^2$

Explanation of Solution

Given:

Photon’s wavelength is equal to the Compton wavelength of a particle

Formula Used:

$\begin{array}{l}E=h\nu \\ \lambda =\frac{h}{mc}\end{array}$

Calculation:

We know that $E=h\nu$

$\begin{array}{l}E=h\nu =hf=\frac{hc}{\lambda }\\ \frac{hc}{\lambda }=\frac{hc}{\frac{h}{m_{0}c}}=m_0{c}^2\end{array}$

Where $m_o$ is the mass of the particle.