1.

How muchradiusare therein stationary orbitals ?

Answer»

Solution :`r_(N) = (n^(2)a_(0))/(Z ) = ((52.9)n^(2))/(z) PM `
wheren = qunatumnumber
orbitalof electron= energylevel
=1.2.3
`a_(0) =52 .9 pm=-0.0529 nm= 5.29 xx 10^(11)m`


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