Elements and Atoms - AP Chemistry

E=hf, or E=hc/λ

h is Planck's constant with a value of 6.62606957 × 10-34 m2 kg / s. Converting 475nm to m gives 4.75x10-7m. c is the speed of light with 3x108m/s. Putting these variables into the second equation you get 4.174x10-19J

de Broglie suggested that all matter has both wave-like and particle-like properties. His
equation states that λ= h/mv , where λ is the wavelength, m is mass, h is Planck’s constant,
and v is velocity. 4 mph is 1.8 m s−1 and Planck’s constant is 6.626e-34 kg m2 s−1 , giving
us a wavelength (λ) 4.6e-36 m ! A very small wavelength, indeed.

Rutherfords experiment focused a beam of alpha particles on a piece of gold foil. The result showed that most of the alpha particles passed through the foil, while a small fraction of particles were significantly deflected. This suggested the presence of a small, dense, positively charged nucleus. Most of the alpha particles passed through the electron clouds of the gold atoms, without impacting the nuclei, while those that did impact the nuclei were deflected by the positive charge of the nucleus.

Rutherford's experiment does not give us any concrete information about neutrons, nor does it allow us to assume that the number of protons and neutrons in the nucleus are equal.

In the Rutherford experiment, a beam of alpha particles was shot through a gold foil, with most of the particles flying straight through and a few scattering at angles greater than . These results suggest that most of the volume of foil was "empty" space, with small concentrations of positive charge, which we now know is the nucleus.

Ernest Rutherford made great advances in current understanding of atomic structure through his gold foil experiment. When firing positively-charged alpha particles at a thin film of gold atoms, most particles were found to pass straight through the film with little to no deflection, indicating that atoms were mostly composed of empty space. A few particles were deflected at large angles, indicating direct collisions with the positively charged nuclei of the atoms.

Recall that J.J. Thomson conducted the cathode ray tube experiment that proved the existence of a small negatively charged particle. Thomson was also able to determine the charge-to-mass ratio of an electron. Robert Millikan is the scientist who conducted the oil drop experiment, from which he was able to find the charge of the electron and deduce its mass using the electron's charge-to-mass ratio. Dalton was not involved in the discovery of the electron.

All the oil drops must have a charge that is a multiple of . If an oil drop is observed to have a charge of , that means the oil drop has three electrons. If an oil drop is observed to have a charge of , that means the oil drop has twelve electrons. If an oil drop is observed to have a charge of , that means the oil drop has two electrons.

To find the number of electrons this oil drop contains, divide the observed charge by the charge of a single electron.

This oil drop contains electrons.

Recall that J.J. Thomson conducted the cathode ray tube experiment that proved the existence of a small negatively charged particle. Thomson was also able to determine the charge-to-mass ratio of an electron. Robert Millikan is the scientist who conducted the oil drop experiment, from which he was able to find the charge of the electron and deduce its mass using the electron's charge-to-mass ratio.

From the cathode ray tube experiment, J.J. Thomson was only able to prove the existence of the negatively charged electron. From the results of the experiment, he also postulated that an atom was made up of a large spherical cloud of positive charge with negatively charged electrons embedded within. This model is also known as the plum pudding model.

Rn is radon, a noble gas, which is not a metal. As a rule, elements to the left are metals, elements to the right are non-metals. The barrier would be the elements that are immideately to the right of the transition metals

As the atomic radius decreases, electrons are drawn closer to the nucleus. Since the electromagnetic force between the positively charged nucleus and negatively charged electrons is a function of distance, the force of attraction, or effective nuclear charge, exerted on each electron will be greater.

To write out the shorthand electronic configuration for Ga, you simply write down the noble gas that that comes before , in this case , then write the rest of the electron configuration as normal.

The atomic number is equal to the number of protons, because no two elements have the same number of protons. It is however possible for different elements to have the same number of electrons (through the loss or gain of electrons) or neutrons (isotopes).

This is the definition of gamma radiation.

In alpha decay, an element loses the equivalent of a helium nucleus, or 2 protons and 2 neutrons. Uranium-238 has an atomic number of 92 and a mass of 238 units, so the product of alpha decay will have an atomic number of 90 and a mass of 234 units. This corresponds to thorium-234.

With each half-life, the amount of Am-242 is halved. 4 days is 96 hours, which is 6 half-lives. For half-life questions use the formula:

Where is the number of half lives.

Thus of Am-242 is left after 4 days.

During alpha decay, an element emits a helium nucleus with 2 neutrons and 2 protons. Thus, the atomic mass of the new element is decreased by four, and the atomic number is decreased by two.

Three subsequent alpha decays result in a new element with an atomic mass of 232 - 3(4) = 220, and a new atomic number of 90 - 3(2) = 84.

Using the periodic table, we find the element with this atomic number is polonium (Po).

In alpha decay, a helium nucleus is emitted, and thus the isotope loses 2 protons and 2 neutrons.

In electron emission, a neutron in the nucleus is converted into a proton and an emitted electron.

In positron emission, a proton in the nucleus is converted into a neutron and an emitted positron.

The given isotope will lose 4 protons and 4 neutrons via alpha emission, gain 1 proton and lose 1 neutron via electron emission, and lose 2 protons and gain 2 neutrons via positron emission. The result is a loss of 5 protons and 8 mass units.

Accounting for the changes in atomic mass and number, we find that the final element is 141-praseodymium.