Quiz 2_Sol

Quiz 2_Sol - De Broglie Wavelength lamda = h/p p=mv 5...

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Quiz 2 answers 1a (Werner) Heisenberg 1b 1c According to the exact same Heisenberg Uncertainty formulation as above, the larger the lifetime, the smaller the uncertainty in energy is going to be, ie, the less broad the “widening” of that energy state (analogous to question on problem set with widening of the Balmer, Paschen, etc lines). Much smaller uncertainty in delta E means a much smaller resonance width, giving you a much higher potential Q factor, for the same frequency, and thus a more precise atomic clock. 2a ANSWER: RUTHERFORD 2b ANSWER:
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Unformatted text preview: De Broglie Wavelength: lamda = h/p; p=mv; 5 points mass = 2*(1.67262158e-27 kg + 1.6749286e-27 kg) = 3.4e-27 kg velocity = 6.3e6 m/s lamda = 1.6e-14 m = 16 fm; 5 points 2c ANSWER: resolution = lamda/2 = 7.9 fm. (5 points) nuclear diameter > resolution = 7.9 fm (10 points) 2.4 fm * m^(1/3) > 7.9 fm m > 35.1 (5 points) Chlorine is the smallest nucleus observable with this "microscope" (5 points) Give 20 points partial credit for the "factor of 2 answers", namely lithium (m > 4.4) and "all are observable" (m > 280) 5 points off for sig figs....
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