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Acceleration & Deflection of Electrons

Background

● Background Overview

The instrument used in this experiment is a cathode ray tube [CRT]. When the filament of this device (also called an electron gun) is heated, electrons (mass Me=9.109×1031M_e = 9.109 \times 10^{-31} kg) are freed and given an acceleration. The acceleration is received when the beam of electrons passes through a region of positive potential with respect to the cathode. Consider the schematic of the tube in Figure 1.

Top) Schematic of the Electron Path in the CRT. Electron enters deflection parallel plates at Point A, exits at Point B, hits screen at Point C having been deflected distance D (in the vertical direction for today’s setup).  Bottom) Electron beam path through the CRT; after being initially generated by the heater, accelerated toward the screen, brightened by the grid voltage and then focused into a narrow beam, the electrons are deflected up or down as they enter the blackened area of the CRT.

Figure 1:Top) Schematic of the Electron Path in the CRT. Electron enters deflection parallel plates at Point A, exits at Point B, hits screen at Point C having been deflected distance DD (in the vertical direction for today’s setup). Bottom) Electron beam path through the CRT; after being initially generated by the heater, accelerated toward the screen, brightened by the grid voltage and then focused into a narrow beam, the electrons are deflected up or down as they enter the blackened area of the CRT.

The electrons receive an amount of kinetic energy given by

K=12Mev2=eVKK = \frac{1}{2} M_e v^{2}= e\cdot V_\text{K}

where ee is the charge of the electron and accelerating voltage VKV_\text{K} (often just VV, but changed here to VKV_\text{K} for the sake of consistency with the apparatus used today) is the potential difference between the heated cathode and accelerating anode. Thus the out-of-the-screen velocity in the Z direction (in line towards the phosphor-coated screen of the CRT) vZv_\text{Z} is given by

vZ=2eVKMev_\text{Z}=\sqrt{\frac{2 e \cdot V_\text{K}}{M_e}}

Then, the accelerated electron beam passes between two parallel plates. If a potential VdV_\text{d} is applied between the plates, the electron beam will be deflected toward the positive plate by a force FE=eEF_E = eE where EE is the electric field between the plates and ee is the electric charge. Thus the acceleration in the Y-direction (a.k.a. transverse direction, see Figure 1) due to the deflection voltage VdV_\text{d} is given by

aY=FEMe=eEMe=eVdMeda_\text{Y} = \frac{F_E}{M_e} = \frac{e E}{M_e} = \frac{e\cdot V_\text{d}}{M_e d}

where dd is the plate separation and MeM_e is the electron mass. Since the electric field between the plates is approximately uniform, the path of the beam will be parabolic in the region between the plates. This is illustrated from Point A to Point B in Figure 1. The path is a case of an object undergoing uniform acceleration just like an object in projectile motion. After the beam leaves this space, it will continue in a straight line with a out-of-the-screen velocity vZv_\text{Z} and a vertical velocity vYv_\text{Y} until it strikes the screen at Point C.

The transverse velocity, i.e. in the Y or ‘deflection’ direction we are using today, is given by

vY=aYtv_\text{Y}=a_\text{Y} t

where tt is the time during which the particle is accelerated as it passes through the plates of length ll. The time t is given by

t=lvZt=\frac{l}{v_\text{Z}}

By combining the above (3), (4), and (5), we obtain the transverse velocity vYv_\text{Y} in terms of measurable quantities, namely

vY=eVdMed(lvZ)v_\text{Y}=\frac{e V_\text{d}}{M_e d}\left(\frac{l}{v_\text{Z}}\right)

Referring to Figure 1, the actual on-screen deflection DD from the axis is vYv_\text{Y} times the time of flight, tt^\prime, over the distance LL from the end of the deflection plates to the screen. Assuming that LlL\gg l, then the time of flight t=L/vZt^\prime = L/v_\text{Z}. Since distance equals velocity ×\times time, we get

D=vYtD=vY(LvZ)D = v_\text{Y} t^\prime \rightarrow D = v_\text{Y}\left(\frac{L}{v_\text{Z}}\right)

As a result of combining everything we’ve seen so far:

D=(lL2dVK)VdD = \left(\frac{l L}{2 d V_\text{K}}\right)V_\text{d}

where we see that deflection is directly proportional to deflection voltage VdV_\text{d}. Usually, the geometric quantities are held constant but it is interesting to take a look at the effect of these factors on the deflection.

With longer parallel plates there is more time for the electric field to act on the electron beam and cause deflection. Also, the closer the spacing between the deflection plates dd, the more intense the field for a given deflection voltage yielding a larger deflection. Finally, the deflection will increase as the accelerating voltage VKV_\text{K} is decreased. This reduces the velocity of the electrons and allows them to spend more time within the deflection field.

The visible spot you see on screen is the result of the energetic electrons releasing their kinetic energy as they impact phosphor atoms. A small portion of that energy is converted to visible light. Most of the energy just heats the screen!

Summing up, the assumptions made in this derivation include:

  1. The geometry of the deflection plates can be considered a ‘parallel-plate’ capacitor configuration. That is, the plates are relatively large and closely spaced. This implies that the electric field is uniform, vertical, and completely contained between the deflection plates.

  2. The distance to the screen is large compared to the horizontal size of the deflection plates (LlL\gg l in Figure 1), so the electron path may be considered to be a straight line from the center of the deflection plate configuration to screen.

  3. The electron beam enters with an initial direction that is parallel to the deflection plates.

● Equipment

Controls and parameters for the CRT of the Complete Properties of Electrons Apparatus.

Figure 2:Controls and parameters for the CRT of the Complete Properties of Electrons Apparatus.

Experimental Procedure

● Procedure Preview & Preliminary Setup

  1. All knobs, switches, displays are numbered and will appear as # in the following steps which correspond to those shown in Figure 2.

  2. Turn on the main power switch 10, and wait until the cathode warms up and a bright spot appears on the screen. If it doesn’t appear after ~30 s, an instructor can be of assistance; it’s likely a deflection or zeroing knob pushing the beam off-screen.

● Experimental Data Collection

  1. FOR CASE 1 as listed on apparatus. Set the CRT voltage selector switch 3 to read the grid voltage VGV_\text{G}. Adjust the grid voltage to the given value using the grid potentiometer 17 whose voltage appears on the top voltmeter display 24.

  2. Now set the CRT voltage selector 3 to read the focus voltage VIV_\text{I}. Adjust the focus voltage to the given value using the focus potentiometer 20 whose voltage also appears on the top voltmeter display 24.

  3. Next, set the CRT voltage selector 3 to read the acceleration or cathode voltage VKV_\text{K}. Set the accel. voltage to the given values using the accel. potentiometer 21 whose voltage also appears on the top voltmeter display 24. (VK, case 1=950V_\text{K, case 1} = 950 V, VK, case 2=1100V_\text{K, case 2} = 1100 V)

  4. Set electrostatic deflection selector switch 8 to X-deflection VXV_\text{X}, and set the X-deflection voltage potentiometers 13 to zero (0.0 V) which appears on the middle voltmeter display 23.

  5. Set electrostatic deflection selector switch 8 to Y-deflection VYV_\text{Y}, and set the Y-deflection voltage potentiometers 12 to zero (0.0 V), again appearing on the middle voltmeter display 23.

  6. Then, use the ZERO X and Y potentiometers 1 to move the bright electron-beam spot to the bottom center of the screen grid (see Figure 3). These knobs do not change the Electrostatic Voltages in 23; if the voltages do change, check you are in fact using the ZERO knobs 1, not 12 or 13. Make sure to view the CRT head-on so as to account for possible misalignment from the effects of parallax (since the grid is horizontally displaced from the glass of the CRT).

Example of CRT grid and 4 mm spacings.

Figure 3:Example of CRT grid and 4 mm spacings.

  1. Ensure the deflection voltage switch 8 is set to read the deflection voltage VYV_\text{Y}. Since we are only deflecting electrons in the Y-direction, we will refer to the deflection voltage as VYV_\text{Y} (i.e. Vd=VYV_\text{d} = V_\text{Y}).

  2. Create a common data table, recording the following CRT data and case setups:

    • dd = spacing between plates listed on the apparatus

    • ll = length of horizontal deflection plates (parallel plates)

    • LL = distance from plates to screen of tube

    • For both cases, record the grid voltages VGV_\text{G}, focus voltages VIV_\text{I}, and accelerating voltages VKV_\text{K}

  3. Create a data table with columns for the deflection voltage VYV_\text{Y} and the deflection position DexperimentalD_\text{experimental} (e.g. 0,4,8,  48mm0, 4, 8,\text{ … }48\,\text{mm}). DexperimentalD_\text{experimental} is based on the CRT grid and not to be calculated. At the start of each case, list the acceleration voltage to gelp clarify which case is which.

  4. Record the initial deflection voltage VYV_\text{Y} and calculate the ratio of the deflection voltage to the accelerating voltage (VY/VKV_\text{Y}/V_\text{K}). Then record the deflection position on the screen (Dexperimental=0mmD_\text{experimental}=0\,\text{mm} at starting point) as well as your best estimate of your uncertainty in deflection position δDexperimental\delta D_\text{experimental}. Then, slowly increase the Y-deflection voltage VYV_\text{Y} using the Y-deflection potentiometer 12. Move the bright spot 4 mm in the Y direction; record the deflection voltage VYV_\text{Y}, ratio VY/VKV_\text{Y}/V_\text{K}, DexperimentalD_\text{experimental} and δDexperimental\delta D_\text{experimental}. Continue this process every 4mm4\,\text{mm} up to 48mm48\,\text{mm} for a total of 13 data points. Double check: The acceleration voltage VKV_\text{K} and the horizontal deflection voltage VXV_\text{X} must be constant during this procedure.

  5. FOR CASE 2, repeat steps 3 to 12 above using Case 2 data listed on the apparatus including an acceleration voltage of 1100 V.

● Experimental Data Analysis

○ Graphical Analysis --- Deflection Voltage (Plot 1)

  1. Plot both cases on the same graph (we want to see direct comparison):

    • Your values of deflection distance vs. deflection voltage (i.e. DexperimentalD_\text{experimental} vs. VYV_\text{Y}).

    • Reminder, DexperimentalD_\text{experimental} is not calculated, but rather just your stated 4 mm increments.

    • Draw a best fit line through each set of data points and show their equations on the single graph.

  2. For each case, determine the relationship between DexperimentalD_\text{experimental} and VYV_\text{Y}. I.e. what is the slope mexperimentalm_\text{experimental} of the best-fit line and the slope’s uncertainty δmexperimental\delta m_\text{experimental} using the LINEST() function (reminders in Plotting in Excel)? Simplified from (8), we can see in the linear form of y=mx+by=mx+b:

    Dexperimental=mexperimentalVYD_\text{experimental}=m_\text{experimental} V_\text{Y}
  3. For each case, calculate the expected slope mexpectedm_\text{expected} as predicted by the rearrangement of equation (8). Since we don’t know the expected DexpectedD_\text{expected} for a given VY,expectedV_\text{Y,expected}, we can instead use info from (8) to get

    mexpected=DexpectedVY,expected=(lL2dVK)m_\text{expected}=\frac{D_\text{expected}}{V_\text{Y,expected}}=\left(\frac{l L}{2 d V_\text{K}}\right)
  4. For each case individually, determine the RMS in the following to use as an estimate of an overall theoretical position uncertainty from a case’s data set:

    a. Use all 13 of your experimental deflection positions DexperimentalD_\text{experimental} and their deflection voltages VYV_\text{Y} as well as the slope mexperimentalm_\text{experimental} of your best-fit line found in Step 15 to calculate the square root of the mean of the square difference between your data and the best fit line (i.e. you’re calculating the root mean square or RMS of the difference between what you claim the deflection position DD is and what the best fit line of your data says the deflection position should be). For each of your 13 data points, you will have a squared difference of (DmVY)2(D - m V_\text{Y})^2.

    b. From those 13 values, you can take the mean, then the square root of that mean value for a resulting single value:

    mean(all 13 squared differences)=mean((DexperimentalmexperimentalVY)2)\sqrt{\text{mean}(\text{all 13 squared differences})} = \sqrt{\text{mean}((D_\text{experimental} - m_\text{experimental} V_\text{Y})^2)}

    where DexperimentalD_\text{experimental} are your claimed 4 mm incremented deflection positions, mexperimentalm_\text{experimental} is your experimental slope, and VYV_\text{Y} are your experimentally determined deflection voltages.

○ Acceleration Voltage Analysis

  1. With your CRT still on, change the deflection voltage VYV_\text{Y} until the beam is in the upper half of the screen.

  2. Using 3, select VKV_\text{K}, and decrease the acceleration voltage with 21 as low as it will go. Now, watch the position on the screen as you increase the accelerating voltage to the maximum value. Did the beam go up or down, why?

○ Graphical Analysis --- VY/VKV_\text{Y}/V_\text{K} (Plot 2)

  1. Graphical Analysis, Plot 2: Plot both cases on the same graph (we want to see direct comparison):

    • Your values of deflection distance vs. deflection voltage to acceleration voltage ratio (i.e. DexperimentalD_\text{experimental} vs. VY/VKV_\text{Y}/V_\text{K}).

    • Draw a best fit line through each set of data points and show their equations on the single graph.

  2. For each case, determine the relationship between DexperimentalD_\text{experimental} and VY/VKV_\text{Y}/V_\text{K}. I.e. what is the slope mexperimental,ratiom_\text{experimental,ratio} of the best-fit line and the slope’s uncertainty δmexperimental,ratio\delta m_\text{experimental,ratio} using the LINEST() function (reminders in Plotting in Excel)? Simplified from (8), we can see (again in the linear form of y=mx+by=mx+b):

    Dexperimental=mexperimental,ratioVYVKD_\text{experimental}=m_\text{experimental,ratio} \frac{V_\text{Y}}{V_\text{K}}
  3. When you are finished with all cases, reset your experimental setup before leaving.

Post-Lab Submission --- Interpretation of Results

● Finalized Spreadsheets

● Post-lab Writeup

The Whiteboard

Overview.

Figure 4:Overview.