Chapter 9, Problem 33P

(a)

To determine

The speed of the electron.

Answer to Problem 33P

The speed of the electron is $0.979c$.

Explanation of Solution

Given info: The kinetic energy of the electron and proton is $2.00\text{MeV}$, the rest energy of the electron and proton are $0.511\text{MeV}$ and $938\text{MeV}$ respectively.

Formula to calculate the total energy of particle is,

$E=\gamma m{c}^2$ (1)

Formula to calculate the total energy of the particle is,

$E=K+m{c}^2$ (2)

Here,

$K$ is the kinetic energy.

$m$ is the mass of particle.

$c$ is the speed of the light in the vacuum.

$E$ is the total energy of the particle.

Equate the equation (1) and equation (2).

$\begin{array}{c}\gamma m{c}^2=K+m{c}^2\\ \gamma =\frac{K}{m{c}^2}+1\end{array}$ (3)

Formula to calculate the Lorentz factor is,

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in equation (3).

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{K}{m{c}^2}+1$

Rearrange the above equation for $v$,

$v=c{\left[1-{\left(\frac{m{c}^2}{K+m{c}^2}\right)}^2\right]}^{1/2}$ (4)

Formula to calculate the rest energy of particle is,

$E_{\text{R}}=m{c}^2$

Substitute $m{c}^2$ for $E_{\text{R}}$ in equation (4).

$v=c{\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]}^{1/2}$ (5)

The speed of the electron is,

$v_{\text{electron}}=c{\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]}^{1/2}$ (6)

Here,

$v_{\text{electron}}$ is the speed of the electron.

Substitute $0.511\text{MeV}$ for $E_{\text{R}}$, and $2.00\text{MeV}$ for $K$ in equation (6) to find the $v_{\text{electron}}$.

$\begin{array}{c}{v}_{\text{electron}}=c{\left[1-{\left(\frac{0.511\text{MeV}}{2.00\text{MeV}+0.511\text{MeV}}\right)}^2\right]}^{1/2}\\ =0.979c\end{array}$

Thus, the speed of the electron is $0.979c$.

Conclusion:

Therefore, the speed of the electron is $0.979c$.

(b)

To determine

The speed of the proton.

Answer to Problem 33P

The speed of the proton is $0.0652c$.

Explanation of Solution

Given info: The kinetic energy of the electron and proton is $2.00\text{MeV}$, the rest energy of the electron and proton are $0.511\text{MeV}$ and $938\text{MeV}$.

From equation (5), the speed of the particle is given as,

$v=c{\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]}^{1/2}$

The speed of the proton is,

$v_{\text{proton}}=c{\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]}^{1/2}$

Here,

$v_{\text{proton}}$ is the speed of the proton.

Substitute $938\text{MeV}$ for $E_{\text{R}}$, and $2.00\text{MeV}$ for $K$ to find the $v_{\text{proton}}$.

$\begin{array}{c}v=c{\left[1-{\left(\frac{938\text{MeV}}{2.00\text{MeV}+938\text{MeV}}\right)}^2\right]}^{1/2}\\ =0.0652c\end{array}$

Thus, the speed of the proton is $0.0652c$.

Conclusion:

Therefore, the speed of the proton is $0.0652c$.

(c)

To determine

The factor by which speed of electron exceed that of the proton.

Answer to Problem 33P

The factor by which speed of electron exceed that of the proton is $15.0$.

Explanation of Solution

Given info: The kinetic energy of the electron and proton is $2.00\text{MeV}$, the rest energy of the electron and proton are $0.511\text{MeV}$ and $938\text{MeV}$.

The ratio of the speed of the electron and proton is,

$\text{ratio}=\frac{v_{\text{electron}}}{v_{\text{proton}}}$

Substitute $0.0625c$ for $v_{\text{proton}}$ and $0.979c$ for $v_{\text{electron}}$.

$\begin{array}{c}\frac{v_{\text{electron}}}{v_{\text{proton}}}=\frac{0.979c}{0.0652c}\\ \frac{v_{\text{electron}}}{v_{\text{proton}}}=15.01\\ v_{\text{electron}}=15.01v_{\text{proton}}\\ v_{\text{electron}}\approx 15v_{\text{proton}}\end{array}$

Thus, the factor by which speed of electron exceed that of the proton is $15.0$.

Conclusion:

Therefore, the factor by which speed of electron exceed that of the proton is $15.0$.

(d)

To determine

The speeds of the electron and proton and the factor by which speed of electron exceed that of the proton.

Answer to Problem 33P

The speeds of the electron and proton are $0.99999997c$ and $0.948c$ respectively and the factor by which speed of electron exceed that of the proton is $1.06$.

Explanation of Solution

Given info: The kinetic energy of the electron and proton is $2000\text{MeV}$, the rest energy of the electron and proton are $0.511\text{MeV}$ and $938\text{MeV}$.

The speed of the electron is,

$v_{\text{electron}}=c{\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]}^{1/2}$ (7)

Substitute $0.511\text{MeV}$ for $E_{\text{R}}$ and $2000\text{MeV}$ for $K$ in equation (7) to find the $v_{\text{electron}}$.

$\begin{array}{c}{v}_{\text{electron}}=c{\left[1-{\left(\frac{0.511\text{MeV}}{2000\text{MeV}+0.511\text{MeV}}\right)}^2\right]}^{1/2}\\ =0.99999997c\end{array}$

The speed of the electron is,

$v_{\text{proton}}=c\left[1-{\left(\frac{E_{\text{R}}}{K+E_{\text{R}}}\right)}^2\right]$ (8)

Substitute $938\text{MeV}$ for $E_{\text{R}}$ and $2000\text{MeV}$ for $K$ in equation (8) to find the $v_{\text{proton}}$.

$\begin{array}{c}{v}_{\text{proton}}=c{\left[1-{\left(\frac{938\text{MeV}}{2000\text{MeV}+938\text{MeV}}\right)}^2\right]}^{1/2}\\ =0.948c\end{array}$

The ratio of the speed of the electron and proton is,

$\text{ratio}=\frac{v_{\text{electron}}}{v_{\text{proton}}}$ (9)

Substitute $0.948c$ for $v_{\text{proton}}$ and $0.99999997c$ for $v_{\text{electron}}$ in equation (9).

$\begin{array}{c}\frac{v_{\text{electron}}}{v_{\text{proton}}}=\frac{0.99999997c}{0.948c}\\ \frac{v_{\text{electron}}}{v_{\text{proton}}}=1.054\\ v_{\text{electron}}=1.0548v_{\text{proton}}\\ v_{\text{electron}}\approx 1.06v_{\text{proton}}\end{array}$

Thus, the speeds of the electron and proton are $0.99999997c$ and $0.948c$ respectively and the factor by which speed of electron exceed that of the proton is $1.06$.

Conclusion:

Therefore, the speeds of the electron and proton are $0.99999997c$ and $0.948c$ respectively and the factor by which speed of electron exceed that of the proton is $1.06$.