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A 10 kg satellite circles earth once every 2h in an orbit having a radius of 8000 km. Assuming that Bohr's angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite. |
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Answer» Solution :`mv_(n) r_(n) `= nh /`2pi` . Here m = 10 kg and `r_(n) 8 xx 10^(6)`m . we have the time period T of the circling satellite as 2h. That is T = 7200 s. THUS the velocity `v_(n) = 2pi r_(n)`/T The quantum number of the orbit of satellite n = `(2pi xx 8 xx 10^(6) m)^(2) xx 10 `/(7200s `xx 6.64 xx 10^(-34)`JS) = 5.3 `xx 10^(45)` Note that the quantum number for the satellite motion is extremely large! In FACT for such large quantum numbers the results of quantization condition tend to those of CLASSICAL PHYSICS. |
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