CHEM-6 Midterm: Chem 6 Dartmouth Spring Exam2S
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Two possible approaches here to find the energy difference and hence the frequency. Either find the energy of both states and subtract, or use one of the formulas that includes the subtraction for you. The energy of the n = 253 and n = 252 states is given by. En = 2. 18x10 18j(z2/n2) where z = 1 for the hydrogen atom. Plug in z = 1, nf = 252, ni = 253, to get = 4. 1x108s 1. The bohr radius is given by r = aon2/z where ao = 0. 529 = 0. 529x10 10 m. Plug in n = 253 and z =1 to get r = 0. 529x10 10m(253)2. That seems a bit small to see unless you"ve got really good vision. But still, mighty big for an atom: en = n2h2/8ml2. Plug in n=1 and 2, plus all the constants. E2 = same thing, but multiplied by 22 instead of 12, so it"s 1. 34x10 17j.
