1.

Findoutthewavelengthof theelectron orbiting in thegroundstateof hydrogenatom .

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")`


Discussion

No Comment Found

Related InterviewSolutions