Notebook Setup:¶

Press the 'Restart Kernel and Run All Cells' button to setup the notebook. Do this only once before starting, since this will delete any data that you have entered in the tables below!

Write your name, section, and today's date in the cell below:¶

Name - Section - Date -

Purpose¶

The purpose of this lab is to determine the charge to mass ratio of the electron.

Information¶

For this lab, we will follow the setup instructions that are given in the manual, given below. Read the sections titled * Introduction * Background information * Principle of the experiment * Safety information * Assembly

With the exception of plugging in the cables, everything has already been assembled. Hook up the cables as shown in Figure 2 in the manual. Then, follow the instructions in the subsection titled "Adjust voltages and currents."

from IPython.display import IFrame
IFrame("./images/em-Apparatus-SE-9629-Manual.pdf", width=800, height=600)

Theory¶

The magnetic force ( $\vec{F}_m$ ) acting on a charged particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field ( $\vec{B}$ ) is given by the Lorentz force $$ \vec{F}_m = q\vec{v} \times \vec{B} \, .$$ Since the electron beam in this experiment is perpendicular to the magnetic field, the equation can be written in scalar form as: $$ F_m = evB $$ where $e$ is the charge of the electron. Since the electrons are moving in a circle, they must be experiencing a centripetal force of magnitude $$ F_c = \frac{mv^2}{r} \, ,$$ where $m$ is the mass of the electron, $v$ is its velocity, and $r$ is the radius of the circular motion. Since the only force acting on the electrons is that caused by the magnetic field, $F_m = F_c$ , these expressions can be equated to give $$ evB = \frac{mv^2}{r} \, , $$ or $$ \frac{e}{m} = \frac{v}{Br} \, .$$ Therefore, in order to determine $e/m$ , it is only necessary to know the velocity of the electrons, the magnetic field produced by the Helmholtz coils, and the radius of the electron beam.

The electrons are accelerated through the accelerating potential $V$ , gaining kinetic energy equal to their charge time the accelerating potential. Therefore $$ eV = \frac{1}{2}mv^2 \, .$$ The velocity of the electrons is therefore $$ v = \left(\frac{2eV}{m}\right)^{1/2} \, .$$ These expressions can be combined to give $$ \frac{e}{m} = \frac{2V}{B^2r2} \, .$$ The magnetic field produced near the axis of a pair of Helmholtz coils is given by the equation $$ B = \frac{N\mu_0 I}{(5/4)^{3/2}a} \, ,$$ where $a$ is the radius of the Helmholtz coils (158 mm), $N$ is the number of turns on each coil (130), $\mu_0$ is the permeability constant ( $4\pi\times 10^{-7}$ T m/A), and $I$ is the current through the Helmholtz coils. Putting this all together gives $$ \frac{e}{m} = \frac{2 V (5/4)^3 a^2}{(N\mu_0 I r)^2} \, . $$

The accepted value of the charge-to-mass ratio of the electron is $$ (e/m)_\mathrm{accepted} = 1.7589\times 10^{11}~\mathrm{C/kg} \, .$$

Equipment Description¶

The $e/m$ Tube — The $e/m$ tube is filled with helium at a pressure of $10^{-2}$ mm Hg, and contains an electron gun and deflection plates. The electron beam leaves a visible trail in the tube, because some of the electrons collide with helium atoms, which are excited and then radiate visible light. The electron gun is shown in Figure 4. The heater heats the cathode, which emits electrons. The electrons are accelerated by a potential applied between the cathode and the anode. The grid is held positive with respect to the cathode and negative with respect to the anode. It helps focus the electron beam.

E/M Tube

Figure 3: $e/m$ Tube.

Electron Gun

Figure 4: Electron Gun.

The Helmholtz Coils — The geometry of Helmholtz coils—the radius of the coils is equal to their separation—provides a highly uniform magnetic field. The Helmholtz coils of the $e/m$ apparatus have a radius and separation of 15.8 cm. Each coil has 130 turns. The magnetic field ( $B$ ) produced by the coils is proportional to the current through the coils.

Procedure¶

Measuring $e/m$ : Method 1¶

Note: there are two knobs that you will need to touch: * Voltage knob on the "Constant Voltage" power supply: Higher voltage means electrons are accelerated more * Current knob on the "Constant Current" power supply: Higher current traveling through the coil gives a higher magnetic field.

Measure the radius of the Helmholtz
Start with the current at 0. For a particular voltage (inital setting is 190 V), increase the current until the beam forms a circle. Adjust the current so that the circle extends equally above and below the scale. This part is important (that the loop formed by the beam is half above and half below the scale).
Carefully measure the radius of the circle formed by the electron beam. Take advantage of the mirror to make sure you are taking the reading head-on (when reading off the distances, align the electron beam with its reflection in the mirror).
Give the values you get for V, I, and r
Give value of V, I, and r Now, as discussed, the equation for the charge-to-mass ratio is given by:

$\frac{e}{m} = \frac{2 V (5/4)^3 a^2}{(N\mu_0 I r)^2} \, .$

Evaluate the equation for e/m. V, I, and r are things you measure (either from the power supplies or the radius measurement). a, N, and $\mu_0$ are constants given earlier.

Give value of e/m, with uncertainty.¶

Measuring e/m: Method 2¶

A more precise value can be determined by combining multiple measurements.

Repeat steps 1-3 for four more values of V (i.e select four more values of V and determine the values of I that give the same value of r).
Input all five sets of values in the table below.

from ipydatagrid import DataGrid
import pandas as pd
import numpy as np

data_table = pd.DataFrame({"Accelerating Voltage (V)": [1,4,9,16,25], \
                                  "Magnet Current (I)": [1,2,3,4,5], \
                                  "Squared Current (I^2)": [1,4,9,16,25], \
                                  "Beam Radius (cm)": [1,1,1,1,1]})

data = DataGrid(data_table, \
                   editable = True, \
                   layout={"height": "140px"})
data.auto_fit_columns = True

print("Data Table")
print(data_table)

---------------------------------------------------------------------------

ModuleNotFoundError                       Traceback (most recent call last)

Cell In[2], line 1
----> 1 from ipydatagrid import DataGrid
      2 import pandas as pd
      3 import numpy as np


ModuleNotFoundError: No module named 'ipydatagrid'

Regrouping the variables in the previously given equation, we get:

$I^2 = \left(\frac{2 (5/4)^3 a^2 m}{(N\mu_0 r)^2 e}\right) V \, .$

Because everything in the parentheses is constant (r is a constant as long as the measurement is performed correctly), if $I^2$ is plotted as a function of V, the slope of the resulting line will be given by $\left(\frac{2 (5/4)^3 a^2 m}{(N\mu_0 r)^2 e}\right)$ . Measuring this slope therefore gives the value of $\left(\frac{2 (5/4)^3 a^2 m}{(N\mu_0 r)^2 e}\right)$ . Regrouping terms allows for solving for $\frac{e}{m}$ .

Analysis¶

Using the cells below, perform a linear fit of the voltage $V$ versus $I^2$ and determine the slope and the uncertainty in the slope. The value of the slope is the value of $V/I^2$ that you will use to determine your experimental value of the electron’s charge-to-mass ratio $e/m$ . Plot a graph of your raw data for $V$ vs $I^2$ along with the linear fit to the data.

First read in the data from your table¶

data.clear_selection() data.select(row1=0, column1=0, row2=4, column2=0) voltage = np.array(data.selected_cell_values)

We are converting the list of values to a numpy array so that we can easily use it in calculations below¶

print("Accelerating Voltage (V):", voltage) data.clear_selection() data.select(row1=0, column1=2, row2=4, column2=2) squared_current = np.array(data.selected_cell_values) print("Squared Current (A^2):", squared_current) data.clear_selection() data.select(row1=0, column1=3, row2=4, column2=3) radius = np.array(data.selected_cell_values) print(r"Beam Radius (cm):", radius) data.clear_selection()

## The line can be fit using numpy's polyfit function, with order 1
## Documentation for polyfit can be found at https://numpy.org/doc/stable/reference/generated/numpy.polyfit.html

fit_params, cov = np.polyfit(x=squared_current, y=voltage, deg=1, cov=True)

slope = fit_params[0]
intercept = fit_params[1]
slope_error = np.sqrt(cov[0,0])

print("Slope (V/A^2):", slope)
print("Intercept (V):", intercept)
print("Slope Error (V/A^2):", slope_error)

## Create your plot here by modifying the dummy code below
## Be sure to label your axes and include units
## Remember you can find matplotlib documentation at https://matplotlib.org/stable/users/index.html

import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots()
ax.plot([1.,2.,3.], [2.9,2.1,1.], 'ro', label = 'data')
ax.plot([1.,2.,3.], [3.,2.,1.], label = 'fit')

ax.set(xlabel='x-axis', ylabel='y-axis', title='Sample plot for demonstration')
ax.legend()
ax.grid()

#fig.savefig("test.png") #This allows you to save the plot to a file
plt.show()

Now calculate the the charge-to-mass ratio of the electron by plugging in the slope. Enter your inferred value of $e/m$ and the associated uncertainty in the cell below.

Enter your response here. Be sure to include units.

Best fit e/m = Uncertainty in e/m = The accepted value of the charge-to-mass ratio of the electron is $$ (e/m)_\mathrm{accepted} = 1.7589\times 10^{11}~\mathrm{C/kg} \, .$$ How does your measured value compare? Comment on your findings. - Enter your response here.

Extra studies¶

Rotate the tube so that it is oriented at an angle with respect to the magnetic field from the Helmholtz coils. As you rotate the tube, observe how the beam deflection is affected. How does the radius of curvature change?. - Give answer here Hold a permanent bar magnet near the tube to show the effect of a magnetic field on the electron beam. Switch between holding the north and south ends of the magnet near the beam. Describe the deflection of the beam. - Give answer here Decrease (or increase if the voltage is already low) the accelerating voltage. Does the curvature caused by the magnet increase or decrease? - Give answer here Does the direction of deflection agree with what you'd expect from the "right hand rule"? How would the beam behave were electrons postively charged? - Give answer here With no magnet and no current in the Helmholtz coils, rotate the tube or the entire apparatus to see the deflection of the beam due to the Earth’s magnetic field. Is the direction of the deflection of the beam as you expect? - Give answer here Why can you see the path of the electron beam (i.e. what causes the glow)? - Give answer here What wavelength is the glow (look up the spectrum of helium and match to the color you see)? - Give answer here When there is no magnetic field, the electron beam seems to stop at the glass tube. Do you think this is because the glass stops the electrons, or do you think it's because the helium gas (the ionization of the gas is what causes the glow) stops at the glass tube? In other words, do you think the beam tunnels through the glass boundary, and we just can't see it because there's no helium outside the tube? Why or why not? - Give answer here How might you test the above? - Give answer here