Millikan Oil Drop Experiment And Advanced Rutherford Experiment

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Click here to erase all data entry on this page E 1-2: Millikan Oil Drop Experiment In the Thomson Cathode Ray Tube Experiment, it was discovered that you can use the deflection of an electron beam in an electric and magnetic field to measure the charge-to-mass ratio ( q / m e ) of an electron. If you then want to know either the charge or the mass of an electron, you need to have a way of measuring one or the other independently. In 1909, Robert Millikan and his graduate student Harvey Fletcher showed that they could make very small oil drops and deposit electrons on these drops (1 to 10 electrons per drop). They would then measure the total charge on the oil drops by deflecting the drops with an electric field. You will get a chance to repeat their experiments and, using the results from the Thomson assignment, be able to experimentally calculate the mass of an electron. 1. To start this activity, click this link for Millikan Oil Drop Experiment . The lab will load in a new tab. Click back to this tab to read further instructions and complete the questions below. 2. What is the purpose of the electron gun in this experiment? The purpose is to create a c How does this source affect the oil droplets in the oil mist chamber? The oil droplets start to become negatively charged when this source begins to provide free electrons in the oil mist chamber. It creates a negative charge. Charge on the oil droplets with electrons.

3. The detector in this experiment is a video camera with a microscopic eyepiece attached to view the oil droplets. Look in the Live Data tab and make sure the camera is on. What do you observe on the video camera screen? I noticed the droplets were falling downwards. Do all the oil drops fall at the same speed? no What force causes the drops to fall? gravity The oil drops fall at their terminal velocity, which is the maximum velocity possible due to frictional forces such as air resistance. The terminal velocity is a function of the radius of the drop. By measuring the terminal velocity ( v t ) of a droplet, the radius ( r ) can be calculated. Then the mass ( m ) of the drop can be calculated from its radius and the density of the oil. Knowing the mass of the oil droplet will allow you to calculate the charge ( q ) on the droplet. IMPORTA N T : Read instructions 4 and 5 before beginning the procedure for 5. 4. Measure the terminal velocity of a drop. Identify a small drop near the top of the window that is falling near the center scale and click the Slow Motion button in the video camera control area in Live Data. Wait until the drop is at a tick mark and start the timer (below the camera display). Let the drop fall for at least two or more tick marks and stop the timer. Do not let the drop fall off the end of the viewing scope. Each tick mark is 0.125 mm. Record the distance and the time in the data table below. 5. Measure the voltage required to stop the fall of the drop. Having measured the terminal velocity, you now need to stop the fall of the drop by applying an electric field between the two voltage plates. This is done by clicking on the buttons on the top or bottom of the Electric Field display below the camera display until the voltage is adjusted such that the drop stops falling. This should be done while in slow motion. When the drop appears to stop, turn the slow motion off and do some final adjustments until the drop has not moved for at least one minute. Record

the voltage, V , indicated on the Electric Field display. Complete the experiment for three drops and record your measurements in the data table. Data Table Drop Distance ( d , in meters) Time ( t , in seconds) Voltage ( V , in volts) 1 0.0002 2.36 187 2 0.0002 2.06 150 3 0.0002 2.18 247 The Millikan Oil Drop Experiment is a classic due to the simplicity of the experimental apparatus and the completeness of the data analysis. The following calculations have reduced very complex equations into simpler ones with several parameters combined into a single constant. Millikan and Fletcher accounted for the force of gravity, the force of the electric field, the density of the oil, the viscosity of the air, the viscosity of the oil, and the air pressure. 0.00025 0.00025 0.00025 6. Calculate the terminal velocity and record the value . Calculate the terminal velocity, v t , in units of using this equation: , where d is the distance the drop fell in meters and t is the elapsed time in seconds. Do not forget that the scale on the viewing scope is in mm (1000 mm = 1 m). Drop # Terminal Velocity ( v t , in m/s) 1 1.05932 2 1.21259 3 1.14678 m s v t = d t 1.05932203 ⋅ 10 - 4 1.21259223 ⋅ 10 - 4 1.14678899 ⋅ 10 - 4

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Each of the equations in instructions 7-10 is shown with and without units. You will find it easier to use the equation without units for your calculations. 7. Calculate the radius (r) of the drop and record the value. With the terminal velocity, you can calculate the radius, in m, of the drop using this equation: Drop # Radius ( r , in meters) 1 9.3049 2 9.9595 3 9.68152 r = ( 9.0407 ⋅ 10 - 5 m ⋅ s ) ⋅ √ v t = ( 9.0407 ⋅ 10 - 5 ⋅ √ v t , without units ) 1 2 1 2 9.30499322 ⋅ 10 - 7 9.95952101 ⋅ 10 - 7 9.68152599 ⋅ 10 - 7

8. Calculate the mass of the drop and record the value. You can use the answer from #7 for the radius ( r ) to calculate the mass of the drop given the density of the oil. The final equation to calculate the mass, in kg, is = Drop # Mass ( m , in kg) 1 2.77064 2 3.39740 3 3.12078 m = V oil ⋅ ρ oil = 4 ⋅ r 3 ⋅ 821 π 3 kg m 3 ( 3439.0 ) ⋅ r 3 kg m 3 ≅ ( 3439.0 ⋅ r 3 , without units ) 2.77064165 ⋅ 10 - 15 3.39740665 ⋅ 10 - 15 3.12078337 ⋅ 10 - 15

9. Since you applied a voltage across the Electric Field to stop the fall of the oil drop, the forces being exerted on the drop must be balanced; that is, the force due to gravity must be the same as the force due to the electric field acting on the electrons stuck to the drop: . Calculate the total charge ( Q tot ) on the oil drop due to the electrons using the equation: where Q ( n ) is the number of electrons on the drop, e is the fundamental electric charge of an electron, m is the mass calculated in #8, and V is the voltage. This answer will provide the total charge on the drop ( Q tot ). The fundamental electric charge of an electron ( e ) is 1.6 ´ 10 -19 C (coulombs). Divide your total charge ( Q tot ) by e and round your answer to the nearest whole number. This is the number of electrons ( Q ( n )) that adhered to your drop. Now divide your total charge ( Q tot ) by Q ( n ) and you will obtain your experimental value for the charge on one electron. Drop # Total charge on drop ( Q tot , in Coulombs) Charge on one electron ( C ) 1 1.45347 1.61497 2 2.22190 1.58707 3 2.42975 1.61983 Q tot = Q(n) ⋅ e = ( 9.810 ⋅ 10 - 2 ) ⋅ = ( 9.81 ⋅ 10 - 2 , without units ) C kg ⋅ J m V m V 1.45347565 ⋅ 10 - 18 1.61497294 ⋅ 10 - 19 2.22190395 ⋅ 10 - 18 1.58707425 ⋅ 10 - 19 2.429752277 ⋅ 10 - 18 1.61983518 ⋅ 10 - 19 What is your average charge for an electron? 1.6010x C 1.6010 ⋅ 10 - 19

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11. Average your results for the charge on one electron. Calculate the percent error by: What is your percent error? 0.625 % % Error = ⋅ 100 your answer - 1.6 ⋅ 10 - 19 1.6 ⋅ 10 - 19 12. You will recall that in the Thomson experiment you were able to calculate the charge-to-mass ratio ( q / m e ) as 1.7 * 10 11 . Using this value for q / m e and your average charge on an electron, calculate the mass of an electron in kg. What is your calculated value for the mass of an electron in kg? 2.72306 kg 2.72306 ⋅ 10 - 8

Click here to erase all data entry on this page E Advanced Rutherford Calculations In this video we'll do an overview of the lab setup for the Rutherford experiment. A key experiment in understanding the nature of atomic structure was completed by Ernest Rutherford in 1911. He set up an experiment that directed a beam of alpha particles (helium nuclei) through a gold foil and then onto a detector screen. He observed that alpha particles were not only emerging in the direction that he expected, but that he could detect alpha particles at all angles, even straight backwards. He described this as “. . . almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” He suggested that the experiment could be understood if almost all of the mass of an atom was concentrated in a small, positively charged central nucleus. In this experiment, you will make observations similar to those of Professor Rutherford. 1. To start this activity, click this link for Rutherford Backscattering . The lab will load in a new tab. Click back to this tab to read further instructions and complete the questions below. Set up the Rutherford experiment by placing the alpha-particle source on the optics table, pointed at the foil holder with a gold foil, and placing the phosphor screen behind the gold foil to detect the alpha particles coming through the foil. 2. Turn on both the alpha source and the phosphor screen. Observe the screen with the alpha- particle beam aiming directly through the foil and into the phosphor screen. 00:52

What do the different signals on the screen mean? The different signals on the screen mean they are alpha particle that are being deflected by the gold foil. These particles are positive. Most of the particles shoot right in the middle while some also hit around the center. 3. Now change the detector to a different location. (If you don't see a signal for a position, you might have to click on the persist button, and wait for a few minutes.) What differences do you see in the signal? When I begin to move around the phosphor screen when in front of the gold foil, I receive the most deflected alpha particles while when on the left or right side of the gold, you get way less when behind it you will get a couple, probably 3 every minute. How many distinct locations can you put the detector in? Seven Eight Five Six Nine How does the signal depend on the angle formed by the source/foil/detector combination? different amounts of alpha particles depending on it locations. If placed in front it will pick up the most while if placed on the side or behind, it will pick up way less. If we move the source from the center to the left or right of the table, the intensity of it will be affected by making it less. the phosphor screen will receive less. When we move around the phosphor screen, it will pick up

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4. Using the Persist button, it is possible to estimate the number of particles that hit the screen as a function of angle. You can do this by counting the particles that hit the screen over a given length of time. Now graph this rate (in particle hits per second) as a function of angle. Attach your graph to this question. IMG_0275.jpeg Choose File What percentage of the particles are being scattered backwards? 3.61 % 5. Assuming that scattering backwards means that the alpha particle hits a nuclei head on, and that the metal foil is 0.001 cm thick, estimate the diameter of a gold nucleus. Note: The atomic diameter of a gold atom is about 2.88 × 10 -10 m Calculations: layers= = Gold nucleus diameter= 5.55*10 m 0.00001 2.88 ⋅ 10 - 10 A nucleus A atom ( ) / layers #scattered sec 100, 000 d nucleus = d atom #scattered sec 100,000 ⋅ layers 5.55 ⋅ 10 - 15

6. If R is the radius of an atomic nucleus, r 0 is the radius of a nucleon, and A is the atomic number, show how R can be approximated by R = r 0 A 1/3 . To find R radius of an atomic nucleus, we will use radius of a nucleon and multiply it by the atomic number of the item used with the exponent of 1/3. The radius of nucleon is R0=1.2 *10^-15 . Then I used the atomic number of 79 for gold with the exponent 1/3. My answer was 5.15*10^-15. 1.2*10^-15(79)^1/3=5.15*10^-15 Edit Insert Formats                        P  7. Given that the value of r 0 is approximately 1.4 fm, calculate the size of the gold nucleus Expected size of gold nucleus: 6.01*10 m 6.01 ⋅ 10 - 15 How well does your previous calculation from your measurements agree with this expected value? My previous calculations are a bit different from the expected value. I probably made some error that lead to not having the correct or similar numbers. Why is your measured value not the same as the expected value? I probably ran into some errors such has when I was timing the screen , I probably did not wait the full time, I might have been off by couple of milliseconds. I might of used the wrong number for the atomic nucleon for question 6.There is many room for error.

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Millikan Oil Drop Experiment And Advanced Rutherford Experiment

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