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the ratio of the radii of n = 10 orbit of hydrogen and Li^(+2) ion is ..... |
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Answer» Solution :`r_(n)=(n^(2))/(Z).a_(0)` where `a_(0)=` Bohr radius `=(H^(2)in_(0))/(pie^(2)m)=0.53Å` For H atom Z=1,n=10 `:.r_(10)=(100)/(1)a_(0)` and For `LI^(+2)` atom Z=3, n=10, R=R `:.R_(10)=(100)/(3)a_(0)"":.(r_(10))/(R_(10))-3` |
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