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Hydrogen atom in ground state is excited by a monochromatic radiation of wavelength lambda = 975 Å then number of spectral lines in resulting spectrum emitted will be ...... |
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Answer» Solution :Energy of incident PHOTON, `E=hf=(hC)/(lambda)` `:.E=(6.625xx10^(-34)xx3xx10^(8))/(975xx10^(-10)xx1.6xx10^(-19))eV` `0.012746xx10^(3)` `=12.75eV` Now `E=-(13.6)/(N^(2))-(-(13.6)/(1^(2)))` `12.75=-(13.6)/(n^(2))+13.6` `:.(13.6)/(n^(2))=13.6-12.75` `:.(13.6)/(n^(2))=0.85eV` `:.n^(2)=(13.6)/(0.85)=16` `:. n=4` `:.` No, of spectral lines, `=(n(n-1))/(2)` `=(4(4-1))/(2)` `=6` |
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