Chapter 4, Problem 5P

A Thomson-type experiment with relativistic electrons. One of the earliest experiments to show that p = γmv (rather than p = mv) was that of Neumann. [G. Neumann, Ann. Physik 45:529 (1914)]. The apparatus shown in Figure P4.5 is identical to Thomson’s except that the source of high-speed electrons is a radioactive radium source and the magnetic field B is arranged to act on the electron over its entire trajectory from source to detector. The combined electric and magnetic fields act as a velocity selector, only passing electrons with speed v, where v = V/Bd (Equation 4.6), while in the region where there is only a magnetic field the electron moves in a circle of radius r, with r given by p = Bre. This latter region (E = 0, B = constant) acts as a momentum selector because electrons with larger momenta have paths with larger radii. (a) Show that the radius of the circle described by the electron is given by r = (l² + y²)/2y. (b) Typical values for the Neumann experiment were d = 2.51 × 10⁻⁴ m, B = 0.0177 T, and l = 0.0247 m. For V = 1060 V, y, the most critical value, was measured to be 0.0024 ± 0.0005 m. Show that these values disagree with the y value calculated from p = mv but agree with the y value calculated from p = γmv within experimental error. (Hint: Find v from Equation 4.6, use mv = Bre or γmv = Bre to find r, and use r to find y.)

Figure P4.5 The Neumann apparatus.

To determine

To show that the radius of the circle described by th electron is given by $r=\frac{\left(l^2+y^2\right)}{2y}$ .

Answer to Problem 5P

It is showed that the radius of the circle described by th electron is given by $r=\frac{\left(l^2+y^2\right)}{2y}$ .

Explanation of Solution

The curved path of the electron is shown in figure 1.

Write the equation for the curved path of the electron.

  ${\left(r-y\right)}^2+l^2=r^2$

Here, $r$ is the radius of the circle described by th electron, $y$ is the vertical distance from the center of the cathode ray tube to the electron impact point on the detector and $l$ is the distance from the cathode ray tube to the photo detector.

Rewrite the above equation.

  $\begin{array}{c}{r}^2-2ry+y^2+l^2=r^2\\ y^2-2ry+l^2=0\end{array}$        (I)

Rewrite the above equation for $r$ .

  $\begin{array}{c}2ry=y^2+l^2\\ r=\frac{l^2+y^2}{2y}\end{array}$

Conclusion:

Therefore, it is showed that the radius of the circle described by th electron is given by $r=\frac{\left(l^2+y^2\right)}{2y}$ .

To determine

To show that the value of $y=0.0024\pm 0.0005\text{ m}$ disagrees with the value of $y$ calculated from $p=mv$ , but agrees with the value of $y$ calculated from $p=\gamma mv$ within experimental error.

Answer to Problem 5P

It is showed that the calculated that the value of $y=0.0024\pm 0.0005\text{ m}$ disagrees with the value of $y$ calculated from $p=mv$ , but agrees with the value of $y$ calculated from $p=\gamma mv$ within experimental error.

Explanation of Solution

Write the equation for the velocity of the electron.

  $v=\frac{V}{Bd}$        (II)

Here, $v$ is the velocity of the electron, $V$ is the applied potential, $B$ is the magnitude of the magnetic field and $d$ is the width of the cathode ray tube.

Write the equation for the momentum of the electron.

  $p=Bre$        (III)

Here, $p$ is the momentum and $e$ is the magnitude of the charge of the electron.

Write the classical expression for the momentum of a particle.

  $p=mv$        (IV)

Here, $m$ is the mass of the electron.

Equate equations (III) and (IV) and rewrite it for $r$ .

  $\begin{array}{c}Bre=mv\\ r=\frac{mv}{Be}\end{array}$        (V)

Write the relativistic equation for the momentum of the particle.

  $p=\gamma mv$        (VI)

Here, $\gamma$ is the Lorentz factor.

Equate equations (III) and (VI) and rewrite it for $r$ .

  $\begin{array}{c}Bre=\gamma mv\\ r=\frac{\gamma mv}{Be}\end{array}$        (VII)

Write the equation for the Lorentz factor.

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

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

Put the above equation in equation (VII).

  $r=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{mv}{Be}$        (VIII)

Write the equation for the root of a quadratic equation $a{y}^2+by+c=0$ .

  $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$        (IX)

Conclusion:

The mass of the electron is $9.11\times {10}^{-31}\text{ kg}$ , the magnitude of the charge of the electron is $1.60\times {10}^{-19}\text{ C}$ and the value of $c$ is $3.0\times {10}^8\text{ m/s}$ .

Substitute $1060\text{ V}$ for $V$ , $0.0177\text{ T}$ for $B$ and $2.51\times {10}^{-4}\text{ m}$ for $d$ in equation (II) to find $v$ .

  $\begin{array}{c}v=\frac{1060\text{ V}}{\left(0.0177\text{ T}\right)\left(2.51\times {10}^{-4}\text{ m}\right)}\\ =2.39\times {10}^8\text{ m/s}\end{array}$

Substitute $9.11\times {10}^{-31}\text{ kg}$ for $m$ , $2.39\times {10}^8\text{ m/s}$ for $v$ , $0.0177\text{ T}$ for $B$ and $1.60\times {10}^{-19}\text{ C}$ for $e$ in equation (V) to find the value of $r$ using equation the classical expression.

  $\begin{array}{c}r=\frac{\left(9.11\times {10}^{-31}\text{ kg}\right)\left(2.39\times {10}^8\text{ m/s}\right)}{\left(0.0177\text{ T}\right)\left(1.60\times {10}^{-19}\text{ C}\right)}\\ =0.0769\text{ m}\end{array}$

Substitute $0.0769\text{ m}$ for $r$ and $0.0247\text{ m}$ for $l$ in equation (I).

  $\begin{array}{c}{y}^2-2\left(0.0769\text{ m}\right)y+{\left(0.0247\text{ m}\right)}^2=0\\ y^2-0.1538y+6.10\times {10}^{-4}=0\end{array}$

Comparison of the above equation with the quadratic equation $a{y}^2+by+c=0$ shows that the value of $a$ is $1$ , the value of $b$ is $-0.1538$ and the value of $c$ is $6.10\times {10}^{-4}$ .

Substitute $-0.1538$ for $b$ , $6.10\times {10}^{-4}$ for $c$ and $1$ for $a$ in equation (IX) to find $y$ .

  $\begin{array}{c}y=\frac{-\left(-0.1538\right)\pm \sqrt{{\left(-0.1538\right)}^2-4\left(1\right)\left(6.10\times {10}^{-4}\right)}}{2\left(1\right)}\\ =0.00408\text{ m or }0.150\text{ m}\end{array}$

The value $0.150\text{ m}$ is wrong since the value of $r$ must be greater than that of $y$ .

The value $0.00408\text{ m}$ is not in agreement with the observed value.

Substitute $9.11\times {10}^{-31}\text{ kg}$ for $m$ , $2.39\times {10}^8\text{ m/s}$ for $v$ , $0.0177\text{ T}$ for $B$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $3.0\times {10}^8\text{ m/s}$ for $c$ in equation (VIII) to find the value of $r$ using equation the relativistic expression.

  $\begin{array}{c}r=\frac{1}{\sqrt{1-\frac{{\left(2.39\times {10}^8\text{ m/s}\right)}^2}{{\left(3.0\times {10}^8\text{ m/s}\right)}^2}}}\frac{\left(9.11\times {10}^{-31}\text{ kg}\right)\left(2.39\times {10}^8\text{ m/s}\right)}{\left(0.0177\text{ T}\right)\left(1.60\times {10}^{-19}\text{ C}\right)}\\ =0.1267\text{ m}\end{array}$

Substitute $0.1267\text{ m}$ for $r$ and $0.0247\text{ m}$ for $l$ in equation (I).

  $\begin{array}{c}{y}^2-2\left(0.1267\text{ m}\right)y+{\left(0.0247\text{ m}\right)}^2=0\\ y^2-0.253y+6.10\times {10}^{-4}=0\end{array}$

Comparison of the above equation with the quadratic equation $a{y}^2+by+c=0$ shows that the value of $a$ is $1$ , the value of $b$ is $-0.253$ and the value of $c$ is $6.10\times {10}^{-4}$ .

Substitute $-0.253$ for $b$ , $6.10\times {10}^{-4}$ for $c$ and $1$ for $a$ in equation (IX) to find $y$ .

  $\begin{array}{c}y=\frac{-\left(-0.253\right)\pm \sqrt{{\left(-0.253\right)}^2-4\left(1\right)\left(6.10\times {10}^{-4}\right)}}{2\left(1\right)}\\ =0.00243\text{ m or }0.251\text{ m}\end{array}$

The value $0.251\text{ m}$ is wrong since the value of $r$ must be greater than that of $y$ .

The value $0.00243\text{ m}$ is in agreement with the observed value $\left(0.0024\pm 0.0005\right)\text{ m}$ .

Therefore, it is showed that the calculated that the value of $y=0.0024\pm 0.0005\text{ m}$ disagrees with the value of $y$ calculated from $p=mv$ , but agrees with the value of $y$ calculated from $p=\gamma mv$ within experimental error.