Atomic Structure

The Rutherford Model of the Atom 1. In 1911 Rutherford proposed the nuclear model of atomic structure. He suggested that an atom consists of a central nucleus (where most of the mass of the atom is concentrated) having a positive charge, surrounded by moving electrons carrying negative charge. Geiger and Marsden carried out an experiment to verify his proposal. The Geiger/Marsden a Particle Scattering Experiment 1. The apparatus is illustrated in the diagram below. | 2. The apparatus was in an evacuated container. The detector was a ZnS screen observed through a low power microscope. Each time an alpha particle hit the screen, a small flash of light was produced. 3. The detector was mounted on a support such that it could be rotated to measure the angular deflection of the alpha particles as they passed through a very thin sheet of gold. They measured the numbers of particles deflected through various angles. 4. It was found that most of ? particles pass through the gold undeflected; only a relatively small number are deflected (scattered). 5. Their results were considered to confirm Rutherford’s model and allowed them to estimate the size of the nucleus (greater than 10-14m) and the size of the atom (greater than10-10m), thus producing the slightly surprising conclusion the most of the space occupied by an atom is empty space! Closest Approach of an Alpha Particle to a Nucleus 1. For a given speed of alpha particle, the closest approach to a nucleus, rmin, will occur when the initial direction of motion of the particle is along the line joining the centers of particle and nucleus. 2. In this case, at the point of closest approach, the speed of the particle is zero. 3. As the particle approaches the nucleus, kinetic energy is being converted to electrical potential energy. K. E. lost = E. P. E. gained| 4. Electrical potential at a distance r from a point charge Q is given by | 5. For a nucleus of atomic number Z, the charge is Ze, where e is the magnitude of the charge on one proton (the same as the magnitude of the charge on an electron). 6. The magnitude of the charge on an alpha particle is 2e 7. Therefore, the energy, w, possessed by an alpha particle placed at distance, rmin, from a charge Ze is given by | 8. So, we have | which gives | Millikan’s Experiment to Measure the Charge on one Electron 1. The diagram below is a very simplified representation of Millikan’s apparatus. | 2. Small drops of oil were allowed to fall into a region between two metal plates, (the top plate had a hole in it). 3. Some of the drops became charged by friction. Further ionization was caused by a beam of x rays. 4. Millikan measured the terminal speed of a drop as it fell through the air, with V = 0. From this he could calculate the radius of the drop (and hence it’s mass). He then applied a voltage, V, to the plates and measured the new terminal speed of the same drop. 5. The change in the terminal speed of the drop was used to calculate the magnitude of the charge on the drop. 6. When many measurements had been done, all the charges were found to be integral multiples of a basic unit of charge, assumed to be the charge on one electron. 7. The value, e, is approximately -1. 6? 10-19 C. 8. A simplified version of Millikan’s experiment can be done by finding the voltage needed to just hold an oil drop stationary between the two plates. 9. Consider a drop having a charge q and mass m. | 10. If the drop is stationary, then the two forces acting on it have equal magnitudes. where E is the field strength. 11. Now, , where d is the distance between the plates, Therefore The Electron Volt (eV) 1. The electron Volt is a unit of work (or energy) much smaller than the Joule. 2. If 1electron moves through a potential difference of 1V then 1eV of work is done. Relation between the Joule and the electron Volt 1. Potential difference is work done per unit charge so, . 1 J is the work done when 1C moves through a p. d. of 1V. 2. The charge on one electron is -1. 6? 10-19 C. 3. Therefore 1eV is the work done when 1 ·6? 10-19C moves through a p. d. of 1V. This means that . 4. To convert energy in J to energy in eV, Experiment to measure the Charge to Mass Ratio of Electrons 1. The method proposed here is similar to that used by J. J. Thomson in 1897. Electrons in an evacuated tube (a â€Å"cathode ray tube†) are sent towards a region of space where there are electric and magnetic fields at 90 ° to each other. If the field strengths have a particular ratio then charged particles can pass through them undeflected. | | | 2. In the following analysis | | | V = voltage accelerating the electrons and producing the electric field between the plates| | v = speed of the electrons| | m = mass of one electron and e = charge on one electron| | E = electric field strength (E = where d = distance between plates)| | B = magnetic flux density| 3. If the electrons pass undeflected (magnitude of electric force equal to magnitude of magnetic force), then it can easily be shown that | 4. To find the speed of the electrons, remember that during acceleration the electrons are losing electric P. E. and gaining K. E. | | E. P. E. lost = K. E. gained| eV = 5. Therefore, | | | 6. Combining equations 1 and 2 to eliminate v gives, | | | | 7. Thus, using his experimental apparatus, Thomson was able to determine the charge-to-mass ratio of the electron. Today, the accepted value of is C kg-1. Atomic Structure The Rutherford Model of the Atom 1. In 1911 Rutherford proposed the nuclear model of atomic structure. He suggested that an atom consists of a central nucleus (where most of the mass of the atom is concentrated) having a positive charge, surrounded by moving electrons carrying negative charge. Geiger and Marsden carried out an experiment to verify his proposal. The Geiger/Marsden a Particle Scattering Experiment 1. The apparatus is illustrated in the diagram below. | 2. The apparatus was in an evacuated container. The detector was a ZnS screen observed through a low power microscope. Each time an alpha particle hit the screen, a small flash of light was produced. 3. The detector was mounted on a support such that it could be rotated to measure the angular deflection of the alpha particles as they passed through a very thin sheet of gold. They measured the numbers of particles deflected through various angles. 4. It was found that most of ? particles pass through the gold undeflected; only a relatively small number are deflected (scattered). 5. Their results were considered to confirm Rutherford’s model and allowed them to estimate the size of the nucleus (greater than 10-14m) and the size of the atom (greater than10-10m), thus producing the slightly surprising conclusion the most of the space occupied by an atom is empty space! Closest Approach of an Alpha Particle to a Nucleus 1. For a given speed of alpha particle, the closest approach to a nucleus, rmin, will occur when the initial direction of motion of the particle is along the line joining the centers of particle and nucleus. 2. In this case, at the point of closest approach, the speed of the particle is zero. 3. As the particle approaches the nucleus, kinetic energy is being converted to electrical potential energy. K. E. lost = E. P. E. gained| 4. Electrical potential at a distance r from a point charge Q is given by | 5. For a nucleus of atomic number Z, the charge is Ze, where e is the magnitude of the charge on one proton (the same as the magnitude of the charge on an electron). 6. The magnitude of the charge on an alpha particle is 2e 7. Therefore, the energy, w, possessed by an alpha particle placed at distance, rmin, from a charge Ze is given by | 8. So, we have | which gives | Millikan’s Experiment to Measure the Charge on one Electron 1. The diagram below is a very simplified representation of Millikan’s apparatus. | 2. Small drops of oil were allowed to fall into a region between two metal plates, (the top plate had a hole in it). 3. Some of the drops became charged by friction. Further ionization was caused by a beam of x rays. 4. Millikan measured the terminal speed of a drop as it fell through the air, with V = 0. From this he could calculate the radius of the drop (and hence it’s mass). He then applied a voltage, V, to the plates and measured the new terminal speed of the same drop. 5. The change in the terminal speed of the drop was used to calculate the magnitude of the charge on the drop. 6. When many measurements had been done, all the charges were found to be integral multiples of a basic unit of charge, assumed to be the charge on one electron. 7. The value, e, is approximately -1. 6? 10-19 C. 8. A simplified version of Millikan’s experiment can be done by finding the voltage needed to just hold an oil drop stationary between the two plates. 9. Consider a drop having a charge q and mass m. | 10. If the drop is stationary, then the two forces acting on it have equal magnitudes. where E is the field strength. 11. Now, , where d is the distance between the plates, Therefore The Electron Volt (eV) 1. The electron Volt is a unit of work (or energy) much smaller than the Joule. 2. If 1electron moves through a potential difference of 1V then 1eV of work is done. Relation between the Joule and the electron Volt 1. Potential difference is work done per unit charge so, . 1 J is the work done when 1C moves through a p. d. of 1V. 2. The charge on one electron is -1. 6? 10-19 C. 3. Therefore 1eV is the work done when 1 ·6? 10-19C moves through a p. d. of 1V. This means that . 4. To convert energy in J to energy in eV, Experiment to measure the Charge to Mass Ratio of Electrons 1. The method proposed here is similar to that used by J. J. Thomson in 1897. Electrons in an evacuated tube (a â€Å"cathode ray tube†) are sent towards a region of space where there are electric and magnetic fields at 90 ° to each other. If the field strengths have a particular ratio then charged particles can pass through them undeflected. | | | 2. In the following analysis | | | V = voltage accelerating the electrons and producing the electric field between the plates| | v = speed of the electrons| | m = mass of one electron and e = charge on one electron| | E = electric field strength (E = where d = distance between plates)| | B = magnetic flux density| 3. If the electrons pass undeflected (magnitude of electric force equal to magnitude of magnetic force), then it can easily be shown that | 4. To find the speed of the electrons, remember that during acceleration the electrons are losing electric P. E. and gaining K. E. | | E. P. E. lost = K. E. gained| eV = 5. Therefore, | | | 6. Combining equations 1 and 2 to eliminate v gives, | | | | 7. Thus, using his experimental apparatus, Thomson was able to determine the charge-to-mass ratio of the electron. Today, the accepted value of is C kg-1.