Rutherford s Gold Foil Experiment.

Transcription

1 Rutherford s Gold Foil Experiment Rob Lucas and Keylink Computers Ltd Introduction This experiment is one of the most famous as it revolutionised our ideas of the atom. Before Rutherford performed this experiment the atom was known to be made of positively and negatively charged particles but the structure was not known. It was generally accepted to be like J. J. Thomson s suggestion of a plum pudding with the positive and negative charges like the fruit in the pudding being interspersed throughout the pudding. It was J. J. Thompson who had discovered the electron in This is known as Thompson s plum pudding model. There was some logic behind this. The charges of the same sign could hardly clump together because of the very strong electrical force that repels charges of the same sign. Ernest Rutherford s experiment (1911) was very simple in its construction, if rather harder to perform. It consisted of a source of alpha particles aimed at a thin gold foil. A screen made of zinc sulphide which scintillated when a particle hit it. That is it created a small light flash. The experimenters used a microscope with a screen attached as a viewer. Question: What is an alpha particle? 1 P a g e

2 An alpha particle source in a lead box with a small hole creates a narrow beam of particles which are aimed at the thin gold foil in the centre of the apparatus. The apparatus (except for the detector) is kept within an evacuated chamber to prevent the alpha particles hitting molecules in the air. Question: What is scintillation? The viewer consisting of the microscope and zinc sulphide screen can be placed anywhere in a circle around the apparatus to view the scintillations. The experimenters performed their observations in the dark after a considerable time taken for their eyes to adjust. They then had to count the scintillations which brought its own difficulties there s an upper limit to just how fast you can count. So the apparatus had to be reconfigured for different angles to make the counting possible. Question: How might the experiment be configured to slow the speed of the necessary counting? What was found was that although the vast majority of particles went straight through the gold foil without changing direction a significant number of them were deflected at angles larger than 90 degrees which simply could not be explained by the plum pudding model. It is relatively easy to calculate the deflection that such a model might produce and the probability of a deflection of 90 degrees or more is vanishingly small (Geiger and Marsden in The Laws of Deflexion of α Particles Through Large Angles. These are the two who actually performed the experiment, it was Rutherford who interpreted the results. You can find this paper on the internet and there s a mistake in the table given on page 7, see if you can spot it). Rutherford put it rather colourfully as It was as if you fired a 15 inch shell at a sheet of tissue paper and it came back to hit you. Rutherford deduced from the experiment that the atom had a positively charged centre that contains most of the mass of the atom, this he called the nucleus. Alpha particles are positively charged so must be repelled by a concentration of positive charges, i.e. the proposed nucleus. As relatively few alpha particles were deflected by the gold foil and very few indeed were bounced back, Rutherford deduced that the nucleus must be very small indeed. 2 P a g e

3 Rutherford actually proposed that the number of particles falling onto a unit area of a screen per second, (call this flux) at an angle of φ is given by this formula: Flux = Qntb 4 cosec 4 (φ/2)/16r 2 Question: Find another way of expressing cosec() in terms of trigonometric functions you are familiar with. Where Q is the number of alpha particles per second emitted, n the number atoms per unit volume in the foil, r is the distance from the foil to the detector, and t is the thickness of the foil. The quantity b is given by: b = 2NeE/mu 2 Ne is the central charge of the atom, m is the mass of the alpha particle and E is its charge and u its velocity. We needn t concern ourselves too much with the detail. The part we are going to concentrate on is that the count of particles in a given time, call this N, is proportional to the fourth power of the cosec of half the angle. That is: N = k cosec 4 (φ/2) You will establish that this relationship holds. The apparatus The apparatus can be seen in the screen shot below. The vacuum pump on the right runs the entire time that you are running the experiment so no further thought needs to be given to it. 3 P a g e

4 The red wheel under the table supporting the evacuated dome turns the detector. To turn it, place the mouse over the red wheel (the colour will change to yellow) and use the mouse wheel. The base of the dome has a scale from which the angle can be read. You may need to zoom in/out using the +/- keys on the number pad to get a clear reading. You can start, stop and reset the counter using the controls on the counter. Place the mouse over the button (which will change colour to yellow) and click with the left mouse button. The method Create a table that has a column for the angle, a column for the count, a column for half the angle, a column for this angle in radians (if your sin calculation is going to need radians) a column for sine of 4 P a g e

5 halt the angle, a column for cosec of half the angle, and finally one for the cosec raised to the power of 4. You may find it convenient to use a spreadsheet program. Phi Count (In 1 min) phi/2 phi/2 rad sin(phi/2) csc(phi/2) (csc(phi/2))^ Create a row for the angle 1 through 10. Using the apparatus fill in the Count values for each angle. Then fill in the remaining columns for each row. To show the relationship we need to plot the second column against the last and verify that this is a straight line. Unfortunately because the range of data is so large this is hard to do. Create a new table which has columns for the logs of the two rows like this: log(n) log csc^ Now we can plot these and verify that what we get is a straight line which shows that Rutherford s relationship between the count and the fourth power of the cosec of half the angle is correct. 5 P a g e

6 Appendix 1 Typical results Phi Count (In 1 min) phi/2 phi/2 rad sin(phi/2) csc(phi/2) (csc(phi/2))^ log(n) log csc^ P a g e

7 logn vs log(cosec^4(phi/2)) P a g e