Electron of energies 10.20 eV and 12.09 eV` can cause radiation to be emitted from hydrogen atoms . Calculate in each case, the principal quantum number of the orbit to which electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.
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`(E)_(n)^(H)=-(13.6)/(n^(2))(eV)`,`therefore (E)_(1)^(H)`=-(13.6)/(1^(2))eV=-13.6 eV (because n=1)`, `therefore (E)_(2)^(H)`=-(13.6)/(2^(2))eV=-3.4 eV (because n=2)`,`therefore (E)_(3)^(H)=-(13.6)/(3^(2))eV=-1.51 eV` (because n=3)Now, `(E)_(2)^(H)-(E)_(1)^(H)`=-3.4 eV +13.6 eV=10.2eV`,`(E)_(3)^(H)-(E)_(1)^(H)=-1.51 eV +13.6 eV=12.9 eV`,Hence electrons will be raised to the state of Principal quantum number 2 and 3 corresponding tp radiation energies 10.2 eV and 12.09 eV respectively.Wave length of radiation emitted for the first is `lambda=(c)/(v)=(hc)/(DeltaE)=(6.62 xx 10^(-34)xx3xx10^(8))/(10.2xx1.6xx10^(-19))=121.7xx10^(-10)` meter =1217 A Similarly, for the other, `lambda=(6.62xx10^(-34)xx3xx10^(8))/(12.09xx1.6xx10^(-19))=1026A`
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