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Consider a hydrogen-like atom whose energy in n^(th) excited state is given by E_(n) = (13.6Z^(2))/(n^(2)) When this excited atom makes transition from excited state to ground state most energetic photons have energyE_(max)= 52.224eV and least energetic photons have energy E_(min) = 1.223eV. find the atomic number of atom and the state of excitation. |
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Answer» Solution :Maximum energy is liberated for transition `E_(n)to1` and minimum energy for `E_(n)toE_(n-1)` HENCE, `(E_(1))/(n^(2))-E_(1)=52.224eV""…(1)` and `(E_(1))/(n^(2))-(E_(1))/((n-1)^(2))=1.224eV""…(2)` Solving above EQUATIONS simultaneously, we GET `E_(1)=-54.4eV` and `n=5` Now `E_(1)=-(13.6Z^(2))/(1^(2))=-54.4eV`. Hence, Z = 2 i.e, gas is HELIUM originally excited to n = 5 energy state. |
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