Question: A good electron thief needs a trap at low energy to entice its prey. A poor electron shepherd will have at least some of its flock dangling out at high energy, consider row 2 and 5 in the periodic table. Why should fluorine, in row 2, is less reactive than rubidium, in row 5?
The electronegativity of fluorine is larger than that of iodine atom and hence fluorine can attract electrons easily from other atoms to form stable chemical bonds. That is why fluorine is more reactive than iodine.
The energy required to remove an electron from the outermost orbit of an atom is called the ionization energy of that atom. If the ionization energy of an atom is low then it is easy to remove an electron from its outermost orbit and hence the atom is more reactive.
The electronegativity of an atom depends both on the atomic number and the distance of the valence electrons from the nucleus of an atom. If the electronegativity of an atom is more, it has more tendency to attract an electron towards itself and hence the atom is more reactive.
The role of lithium and rubidium is to "give up" an electron. Hence, they are more reactive if they have low ionization energy. The valence electrons in lithium are closer to the nucleus compared that in rubidium atom. Therefore, the ionization energy of rubidium is less compared to that of lithium. That means, it is harder to remove an electron from lithium compared to that of rubidium. Hence rubidium is more reactive than lithium.
The role of fluorine and iodine is to "seize" an electron. Hence they are more reactive if they are more electronegative. The size of the fluorine atom is much smaller than the iodine atom. Thus, the outermost electron in iodine is far from the nucleus compared to that in the Fluorine atom. Therefore electronegativity of fluorine is larger than that of the iodine atom and hence fluorine can attract electrons easily from other atoms to form stable chemical bonds. That is why fluorine is more reactive than iodine.
The wave functions for the ground and first excited states of a simple harmonic oscillator are and . Suppose you have two particles occupying these two states.
(a) If distinguishable, an acceptable wave function would be . Calculate the probability that both particles would be on the positive side of the origin and divide by the total probability for both being found over all values of , . (This kind of normalizing-as-we-go will streamline things.)
(b) Suppose now that the particles are indistinguishable. Using the symbol to reduce your work. calculate the same probability ratio, but assuming that their multiparticle wave function is either symmetric or antisymmetric. Comment on your results.
The spin-orbit interaction splits the hydrogen 4f state into many.
(a)Identity these states and rank them in order of increasing energy.
(b)If a weak external magnetic field were now introduced (weak enough that it does not disturb the spin-orbit coupling). Into how many different energies would each of these states be split?
Question: In classical electromagnetism, the simplest magnetic dipole is a circular current loop, which behaves in a magnetic field just as an electric dipole does in an electric field. Both experience torques and thus have orientation energies -p.E and (a) The designation "orientation energy" can be misleading. Of the four cases shown in Figure 8.4 in which would work have to be done to move the dipole horizontally without reorienting it? Briefly explain. (b) In the magnetic case, using B and u for the magnitudes of the field and the dipole moment, respectively, how much work would be required to move the dipole a distance dx to the left? (c) Having shown that a rate of change of the "orientation energy'' can give a force, now consider equation (8-4). Assuming that B and are general, write in component form. Then, noting that is not a function of position, take the negative gradient. (d) Now referring to the specific magnetic field pictured in Figure 8.3 which term of your part (c) result can be discarded immediately? (e) Assuming that and vary periodically at a high rate due to precession about the z-axis what else may be discarded as averaging to 0? (f) Finally, argue that what you have left reduces to equation (8-5).
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