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Calculate the wavelength for the emission transition if it starts from the orbit having radius `1.3225 nm` ends at `211.6 p m`. Name the series to which this transition belongs and the region of the spectrum. |
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Answer» Radius of nth orbit of H-like particles `=(0.529n^(2))/Z Å=(52.9n^(2))/Z p m` `r_(1)=1.3225 nm=1322.5 p m =52.9 n_(1)^(2)` `r_(2)=211.6 p m=(52.9n_(2)^(2))/Z` `:. r_(1)/r_(2)=1322.5/211.6=n_(1)^(2)/n_(2)^(2)` or `n_(1)^(2)/n_(2)^(2)=6.25` or `n_(1)/n_(2)=2.5` `:.` If `n_(2)=2, n_(1)=5`. Thus the transition is from `5th` orbit to `2nd` orbit. It belongs to Balmer series. `bar(v)=1.097xx10^(7)m^(-1)(1/2^(2)-1/5^(2))` `=1.097xx21/100xx10^(7) m^(-1)` or `lambda=1/bar(v)=100/(1.097xx21xx10^(7))m` `=434xx10^(-9)m=434 nm` It lies in the visible region. |
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