Chapter 25, Problem 33E

(a)

To determine

Ratio of total energy of moving electron to rest energy.

Answer to Problem 33E

The ratio of total energy to rest energy of electron is 1.12.

Explanation of Solution

Electron initially at rest is accelerated through potential difference $60000\text{V}$.

Write the expression of ratio of total energy to rest energy

  $f=\frac{\left(T+m{c}^2\right)}{\left(m{c}^2\right)}$        (I)

Here, $T$ is kinetic energy of moving electron and $m{c}^2$ is rest energy of electron.

Since the electron is accelerated through potential difference of $60000\text{V}$ therefore

  $T=60000\text{eV}$

Conclusion:

Substitute $60000\text{eV}$ for $T$ and $0.511\text{MeV}$ for $m{c}^2$ in expression (I)

  $\begin{array}{c}f=\frac{\left(\left(60000\text{eV}\right)+\left(0.511\text{MeV}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\right)\right)}{\left(\left(0.511\text{MeV}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\right)\right)}\\ =1.12\end{array}$

Thus, ratio of total energy to rest energy of electron is 1.12.

(b)

To determine

Final speed of electron.

Answer to Problem 33E

The final speed of electron is $1.35\times {10}^8\text{m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Write the expression of total energy $E$ of moving electron

  $E=\frac{m{c}^2}{{\left(1-\frac{u^2}{c^2}\right)}^{\frac{1}{2}}}$        (II)

Here, $u$ is final speed of electron.

Since the ratio of total energy to rest energy is $1.12$ therefore

  $E=\left(1.12\right)m{c}^2$

Conclusion:

Substitute $\left(1.12\right)m{c}^2$ for $E$ in expression (II)

  $\left(1.12\right)m{c}^2=\frac{m{c}^2}{{\left(1-\frac{u^2}{c^2}\right)}^{\frac{1}{2}}}$

Simplify above expression for $u$

  $u={\left(1-\frac{1}{{\left(1.12\right)}^2}\right)}^{\frac{1}{2}}c$

Substitute $3\times {10}^8\text{m}\cdot {\text{s}}^{-1}$ for $c$ in above expression

  $\begin{array}{c}u={\left(1-\frac{1}{{\left(1.12\right)}^2}\right)}^{\frac{1}{2}}\left(3\times {10}^8\text{m}\cdot {\text{s}}^{-1}\right)\\ =1.35\times {10}^8\text{m}\cdot {\text{s}}^{-1}\end{array}$

Thus, the final speed of electron is $1.35\times {10}^8\text{m}\cdot {\text{s}}^{-1}$.

(c)

To determine

Final speed of electron as governed by nonrelativistic mechanics.

Answer to Problem 33E

The final speed of electron according to nonrelativistic mechanics is $1.45\times {10}^8\text{m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Write the expression of final speed $v$ of electron according to nonrelativistic mechanics

  $v={\left(\frac{2T}{m}\right)}^{\frac{1}{2}}$        (III)

Here, $m$ is mass of electron.

The value of kinetic energy of electron is:

  $\begin{array}{c}T=60000\text{eV}\\ \text{=60000}\left(1.6\times {10}^{-19}\right)\text{J}\end{array}$

Conclusion:

Substitute $\text{60000}\left(1.6\times {10}^{-19}\right)\text{J}$ for $T$ and $9.11\times {10}^{-31}\text{kg}$ for $m$ in expression (III)

  $\begin{array}{c}v={\left(\frac{2\left(\text{60000}\left(1.6\times {10}^{-19}\right)\text{J}\right)}{\left(9.11\times {10}^{-31}\text{kg}\right)}\right)}^{\frac{1}{2}}\\ =1.45\times {10}^8\text{m}\cdot {\text{s}}^{-1}\end{array}$

Thus, electron has final speed $1.45\times {10}^8\text{m}\cdot {\text{s}}^{-1}$ according to nonrelativistic mechanics.