1.

The angular speed of the electron in the n^(th) Bohr orbit of the hydrogen atom is proportional to......

Answer»

N
`n^(3)`
`(1)/(n)`
`(1)/(n^(3))`

Solution :Angular momentum `mvr=mr^(2)OMEGA=(nh)/(2pi)`
Angular velocity `omega=(nh)/(2pimr^(2))`
`:.r^(2)=(n^(4)h^(4)epsi_(0)^(2))/(pi^(2)m^(2)E^(4))`
`omega=(pime^(4))/(2h^(3)in_(0)^(2)).(1)/(n^(3))`
`:.omegaprop(1)/(n^(3))`


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