For deuterium $$12H$$ or $$12D$$, the value of Rydberg constant is $$109707$$ $$cm-$$. (This$$ value reflects a refinement of simple Bohr theory, wherein the orbital radii and Rydberg constant do not depend on electron mass but on the so called reduced mass. The latter mass, in turn, varies slightly with the mass of the $$nucleus). Find the shortest wavelength in the absorption spectrum of deuterium

Question

For deuterium  $$12H$$ or $$12D$$, the value of Rydberg constant is $$109707$$ $$cm-$$. (This$$ value reflects a refinement of simple Bohr theory, wherein the orbital radii and Rydberg constant do not depend on electron mass but on the so called reduced mass. The latter mass, in turn, varies slightly with the mass of the $$nucleus). Find the shortest wavelength in the absorption spectrum of deuterium

Infinity Learn · Accepted Answer

(i) R for H $$12$$ $$=$$ $$109707$$ $$cm-$$ The shortest wavelength transition corresponds to highest frequency and hence to the highest energy $$($$ ∵E $$=$$ $$hu)$$. So, the transitionwill be from ground state (lowest$$ energy $$state) for which n $$=$$ $$1$$ to the highest state for which n $$=$$∞.Hence,      u¯$$=RZ21n12$$−$$1n22cm$$−                  $$=109707$$×$$12112$$−$$1($$∞$$)2cm$$−or          $$1$$λ$$=109707(1$$−$$0)cm$$−$$=109707cm$$−∴ λ$$=1109707cm$$−$$=9.1152$$×$$10$$−$$6cm=9.1152$$×$$10$$−$$6cm$$×$$107nm1cmor$$                 λ$$=91.152$$ nm    Ans.

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