Chapter 26, Problem 38AP

(a)

To determine

The speed of proton that has the same kinetic energy of electron.

Answer to Problem 38AP

The speed of proton that has the same kinetic energy of electron is $7.08\times {10}^3\text{km/s}$.

Explanation of Solution

Given Info:

The rest energy of electron is $938.3\text{MeV}$.

The rest energy of proton is $0.511\text{MeV}$.

The speed of electron is $0.750c$.

The speed of light is $3.00\times {10}^8\text{m/s}$.

The formula used to calculate the gamma factor of electron is,

${\gamma }_e=\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}$

• ${\gamma }_e$ is the gamma factor of electron
• $v_e$ is the velocity of electron
• $c$ is the speed of light

The formula used to calculate the gamma factor of proton is,

${\gamma }_p=\frac{1}{\sqrt{1-{\left(\frac{v_p}{c}\right)}^2}}$

• ${\gamma }_p$ is the gamma factor of proton
• $v_p$ is the velocity of proton

The formula used to calculate the kinetic energy of electron is,

$\begin{array}{c}K.E=\left({\gamma }_e-1\right)E_R\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}-1\right)E_{R_e}\end{array}$ I

• $K.E$ is the kinetic energy
• $v_e$ is the speed of electron
• $E_{R_e}$ is the rest energy

The formula used to calculate the kinetic energy of proton is,

$\begin{array}{c}K.E=\left({\gamma }_p-1\right)E_R\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_p}{c}\right)}^2}}-1\right)E_{R_p}\end{array}$ II

• $v_{{}_p}$ is the speed of electron
• $E_{R_p}$ is the rest energy

Since, the kinetic energies are same, equate the equations I and II .

$\left(\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}-1\right)E_{R_e}=\left(\frac{1}{\sqrt{1-{\left(\frac{v_p}{c}\right)}^2}}-1\right)E_{R_p}$

 Rearrange the equation to find $v_{{}_p}$.

$v_p=c\sqrt{1-{\left(\frac{E_{R_p}}{E_{R_e}}\right)}^2\frac{1}{{\left(\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}-1+\frac{E_{R_p}}{E_{R_e}}\right)}^2}}$

Substitute $938.3\text{MeV}$ for $E_{R_p}$, $0.511\text{MeV}$ for $E_{R_e}$ and $0.750c$ for $v_e$ to find $v_p$.

$\begin{array}{c}{v}_p=c\sqrt{1-{\left(\frac{938.3\text{MeV}}{0.511\text{MeV}}\right)}^2\frac{1}{{\left(\frac{1}{\sqrt{1-{\left(\frac{0.750c}{c}\right)}^2}}-1+\frac{938.3\text{MeV}}{0.511\text{MeV}}\right)}^2}}\\ =0.0236c\\ =\left(0.0236c\right)\left(\frac{3.00\times {10}^8\text{m/s}}{1c}\right)\left(\frac{1\text{km}}{{10}^3\text{m}}\right)\\ =7.08\times {10}^3\text{km/s}\end{array}$

Thus, the speed of proton that has the same kinetic energy of electron is $7.08\times {10}^3\text{km/s}$.

Conclusion:

The speed of proton that has the same kinetic energy of electron is $7.08\times {10}^3\text{km/s}$.

(b)

To determine

The speed of proton that has the same momentum as the electron.

Answer to Problem 38AP

The speed of proton that has the same momentum as the electron is $185\text{km/s}$

Explanation of Solution

Given Info:

The mass of electron is $9.11\times {10}^{-31}\text{kg}$.

The mass of proton is $1.67\times {10}^{-27}\text{kg}$.

The formula used to calculate the momentum of electron is,

$\begin{array}{c}P={\gamma }_e{m}_e{v}_e\\ =\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}{m}_e{v}_e\end{array}$ I

• $P$ is the momentum
• $m_e$ is the mass of electron

The formula used to calculate the momentum of electron is,

$\begin{array}{c}P={\gamma }_e{m}_e{v}_e\\ ==\frac{1}{\sqrt{1-{\left(\frac{v_p}{c}\right)}^2}}{m}_p{v}_p\end{array}$ II

• $m_p$ is the mass of proton

Since, the kinetic energies are same, equate the equations I and II .

$\frac{1}{\sqrt{1-{\left(\frac{v_e}{c}\right)}^2}}{m}_e{v}_e=\frac{1}{\sqrt{1-{\left(\frac{v_p}{c}\right)}^2}}{m}_p{v}_p$

 Rearrange the equation to find $v_{{}_p}$.

$v_p=\sqrt{\frac{{\left(m_e{v}_e\right)}^2}{{\left(m_p\right)}^2\left(1-{\left(\frac{v_e}{c}\right)}^2\right)+{\left(\frac{m_e{v}_e}{c}\right)}^2}}$

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $1.67\times {10}^{-27}\text{kg}$ for $m_p$ and $0.750c$ for $v_e$ to find $v_p$.

$\begin{array}{c}{v}_p=\sqrt{\frac{{\left(\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.750c\right)\right)}^2}{{\left(1.67\times {10}^{-27}\text{kg}\right)}^2\left(1-{\left(\frac{0.750c}{c}\right)}^2\right)+{\left(\frac{\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.750c\right)}{c}\right)}^2}}\\ =\sqrt{\frac{{\left(9.11\times {10}^{-31}\text{kg}\right)}^2\left({0.750}^2\right){\left(3.00\times {10}^8\text{m/s}\right)}^2}{{\left(1.67\times {10}^{-27}\text{kg}\right)}^2\left(1-{0.750}^2\right)+{\left(9.11\times {10}^{-31}\text{kg}\right)}^2{\left(0.750\right)}^2}}\\ =185\times {10}^3\text{m/s}\\ \text{=}\left(185\times {10}^3\text{m/s}\right)\left(\frac{1\text{km}}{{10}^3\text{m}}\right)\\ \text{=185}\text{km/s}\end{array}$

Thus, the speed of proton that has the same momentum as the electron is $185\text{km/s}$.

Conclusion:

The speed of proton that has the same momentum as the electron is $185\text{km/s}$