To investigate the claim that lower ℓ implies lower energy, consider a simple case: lithium, which has two n= 1 electrons and a lone n = 2 valence electron.
(a) First find the approximate orbit radius, in terms of ao, of an n = 1 electron orbiting three protons. (Refer to Section 7.8.)
(b) Assuming the n - 1 electrons shield/cancel out two of the protons in lithium's nucleus, find the orbit radius of an n = 2 electron orbiting a net charge of just +e, (c) Argue that lithium's valence electron should certainly have lower energy in a 2s state than in a 2p state. (Refer to Figure 7.15.)
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