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Findoutthewavelengthof theelectron orbiting in thegroundstateof hydrogenatom . |
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Answer» SOLUTION :As perquantum concept, for a stable ntheorbitof HYDROGEN atomthe total length oforbital path (i.e., the circumferenceof the orbit) is equal to n times the de - BROGLIE wavelenghtof electron. Thus , `2 pi rn = n lambda_("electron")` For an electron orbitingin the ground stateof hydrogenatom , n =1 . HENCE , we have ` 2 pi r_(1) = lambda_("electron")` |
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