InterviewSolution
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InterviewSolution
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A hydrogen - like atom (described by the Bohr model) is observed to emit six wavelength , originating from all possible transitions between `- 0.85 eV and -0.544 eV` (including both these values ) (a) Find the atomic number of the atom (b) Calculate the smallest wavelength emitted in these transitions . (Take `hc = 1240 eV - nm`, ground state energy of hydrogen atom `= 13.6 eV)` |
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Answer» Correct Answer - (i) `Z = 3`, (ii) `4052.3 nm` (i) total 6 lines are emitted. Therefore `(n(n-1))/(2)=6 implies n=4` So, transition is taking place between `m^(th)` energy state and `(m+3)^(th)` energy state. `E_(m)= -0.85 eV` `implies -13.6 (Z^(2)/m^(2))= -0.85" "implies Z/m=0.25` …(i) Similarly `E_(m+3)= -0.544 eV` `implies -13.6 z^(2)/((m+3)^(2))= -0.544 implies z/((m+3))=0.2` ...(ii) Solving equation (i) and (ii) for z and m. We get `m=12` and `z=3` (ii) Smallest wavelength corresponds to maximum difference of energies which is obviously `E_(m+3)-E_(m)`. `:. DeltaE_("max")= -0.544-(-0.85)=0.306 eV` `:. lambda_("min")=(hc)/(DeltaE_("max"))=1240/0.306=4052.3 nm`. |
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