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# test-3 - Test 3 Solutions 1(a The ground state of the...

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Test # 3 Solutions 1(a) The ground state of the Hydrogen atom ( n = 1) is E 1 = - 13 . 6eV. Thus, we require exactly 13 . 6eV to ionize the Hydrogen atom in its ground state. 1(b) With E initial = 0 eV and E final = - 3 . 4 eV, we have E = hf = E initial - E final = 0 - ( - 3 . 4) = 3 . 4 eV. Then, using the fact that h = 4 . 14 × 10 - 15 eV s, and that f = c/λ where c = 3 . 0 × 10 8 m s - 1 is the speed of light, we get c λ = E initial - E final = 3 . 4 eV λ = hc E = 4 . 14 × 10 - 15 eV s × 3 . 0 × 10 8 m s - 1 3 . 4eV = 3 . 65 × 10 - 7 m Thus, the emitted photon has a wavelength of nearly 3 . 65 × 10 - 7 m. We may also use the alternative approach, where we use the formula 1 λ = R ( 1 n 2 - 1 m 2 ) In the above expression, R = 1 . 097 × 10 7 m - 1 . However, we now need to be careful, since we are not starting off with the Hydrogen atom. Instead, we are starting out with a single proton, to which an electron is brought so that they together form a Hydrogen atom in the first excited state ( n = 2). But initially, we were not in the Hydrogen atom at all, so the proper value for m in this setup would be m = for which (1 /m ) = (1 /m 2 ) = 0. You can check for yourself that if we plug these values above, we get the same answer,

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test-3 - Test 3 Solutions 1(a The ground state of the...

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