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Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different transitions are possible in the resulting emission spectrum? Find the longest wavelength amongst these. (lonisation energy of hydrogen atom in its ground state is 13.6 eV and take h=6.6xx10^(-34)Js) Data : Wavelength of incident radiation = 970.6 Å = 970.6 X 10^(-10)m Ionisation energy of hydrogen atom in its ground state = 13.6 eV (i) Number of possible transitions = ? (ii) Longest wavelength emitted = ? |
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Answer» Solution :(i) Energy of the excited state, `E=(HC)/(LAMBDA)=(6.6xx10^(-34)xx3xx10^(8))/(970.6xx10^(-10)xx1.6xx10^(-19))=12.75eV` `E_(n)=-13.6+12.75=0.85eV` `E_(n)=(-13.6)/(n^(2))` or `n^(2)=(-13.6)/(E_(n))=(-13.6)/(-0.85)=16` (ii) or n=4 (ii) The number of POSSIBLE transitions in going to the lower state and HENCE the number of different wavelengths in the spectrum will be six (shown in the FIGURE). The longest wavelength corresponds to minimum energy difference. (i.e., for transition 4 `to `3) `E_(3)=(-13.6)/(3^(2))+-1.151eV` `(hc)/(lambda_("max"))=E_(4)-E_(3)` or `lambda_("max")=(6.6xx10^(-34)xx3xx10^(8))/((1.51xx0.85)xx1.6xx10^(-19))` `=18750Åm` `lambda_("max")=18750Å` 
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