posted by jenish .
Lithium is the atom which has three protons in its nucleus and three electrons circling the nucleus.(it usuallly has 3 nuetrons as well,but they r not important in this problem.)Two of the electrons circle the nucleus at a distance 1.8* 10^-11 m (app.)The third electron is farther away at about 1.4 * 10^-10 m from the nucleus and is called the valence electron.
a)Treating the nucleus as a point of charge of value +3e find the force on the valence electron.
b)The electrical force is what supplies the centripetal acceleration (v^2/r) necessary for the electrons circular motion.With what speed does the valence electron circle the nucleus?
c)Find the ionisation energy i.e the energy needed by an electron to completely escape its nuclues, for the lithium atom's valence electron.The ionisation energy is the negative of the electrons total energy E=1/2mv^2 + PE. the PE is that of an electron starting from infinity and being brought to the distance given above from the point charge nucleus.
On a), don't forget that the force on the valence electron includes the attractive force of the nuclues, and the repulsive forces from the inner electrons. So, for a gross simplification, assume a net charge of +1e at the nucleus. Yes, this is a gross simplifiation, but assuming the electrons are circling the nucleus is just as big assumption. Use coulombs law to find the force between the nucleus (+e), and the valence electron (-e).
b) F=mv^2/r solve for v.
c) to find PE, assuming you know calculus, PE=int Force(x)*dx from inf to r. If you don't know calculus, it should work out to ke^2/r .
hey Mr. Bob
I was just wondering why did u take the charge of nucleus +e instead of 3e which is unclear to me..also we gotto find the total force on valence electron ie from both nucleus and the two electrons..am i correct....if not can u help me out..