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How muchradiusare therein stationary orbitals ? |
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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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