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What is the wavelength of light for the least energetic photoa emitted in the Lyman series of the hydrogen spectrum. (take hc = 1240 eV nm): |
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Answer» 102nm `n_(1)=1 and n_(2)=2` then `E_(1)=-13.6eV and E_(2)=-3.4eV` `E=E_(2)-E_(1)=-3.4 +13.6=10.2eV` Now `E=hv=(HC)/(lambda)` `lambda=(hc)/(E)=(1240eV)/(10.2eV)NM` `lambda=122nm`. |
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