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The wavelength of the first line of Balmer series in hydrogen atom is `6562.8Å`. Calculate ionisation potential of hydrogen and also, the wavelength of first line of Lyman series. |
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Answer» Here, `lambda_1=6562.8Å=6562.8xx10^(-10)m` `:. 1/(lambda_1)=R(1/(2^2)-1/(3^2))=Rxx5/36` `R=36/(5lambda_1)=36/(5xx6562.8xx10^(-10))` `R=1.0971xx10^7m^-1` The energy of electron in nth energy state `E_n=R(hc)/(n^2)` `:.` Ionisation energy of Hydrongen atom `E=E_1-E_(oo)=Rhc-0` `=1.0971xx10^7xx6.6xx10^(-34)xx3xx10^8` `=21.788xx10^(-19)J` `=(21.78xx10^(-19))/(1.6xx10^(-19))eV=13.6eV` `:.` Ionisation potential of hydrogen atom `=13.6V` The wavelength of first line of Lyman series is given by `1/lambda=R(1/(1^2)-1/(2^2))=3/4R` `lambda=4/(3R)=4/(3xx1.0971xx10^7)` `=1.2153xx10^-7m=1215.3Å` |
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