548 Molecules and SolidsP43.2We are toldK CleVKCl++→ ++−07.andCl eCleV+→ +−−36.orClCl eeV→+−..By substitution,K CleVKCl+eeV−K eVK+e+→43.or the ionization energy of potassium is . eV.P43.3(a)Minimum energy of the molecule is found fromdUdrArBr=−+=1260137yielding rAB0162=LNMOQP.(b)EUUAABBBABArrr=−=−−LNMMOQPP−LNMOQP==∞=00421412422This is also the equal to the binding energy, the amount of energy given up by the twoatoms as they come together to form a molecule.(c)r0120601120124 101488 10742 10742=×⋅LNMMOQPP=×=−−−..eV meV mm pm126ejE==−−40124 10446602120...eV meV meV612*P43.4(a)We add the reactions KeVKe+434.andI eIeV+→+306.to obtainK IKIeV+ +−434 306af.The activation energy is 128eV.(b)dUdrrr=∈−FHGIKJ+FHGIKJLNMMOQ
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .