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Find the number of waves made by a Bohr electron in one complete revolution in the 3rd orbit. |
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Answer» Solution :The NUMBER of WAVES in any orbit `= ("CIRCUMFERENCE")/("wavelength")=(2pir)/(LAMBDA)=2pir xx(mv)/(h)` The angular momentum of 3rd orbit is `(3h)/(2pi)` Number of waves `=(2pi)/(h)xx(3h)/(2pi)=3` |
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