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The Rydberg constant of H_2 atom is 10967700 m^(-1). Calculate the short and long wavelength limits in its Lyman series : |
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Answer» 602 Å, 906Å `1/(lambda_("short"))=R [1/1^(2)-(1)/((oo)^(2))]=R=10967700` `lambda_("short")=(1)/(10967700)=0.911 XX 10^(-7) m=911Å` For long wavelength, n=2 `1/(lambda_("short"))=R [1/1^(2)-(1)/((oo)^(2))]=R=10967700` `lambda_("short")=(1)/(10967700)=0.911xx 10^(-7)m=911Å` For long wavelength, n=2 `1/lambda_("long")=R [1/1^(2)-1/2^(2)]=3/4R` `lambda_("long")=1212 Å` |
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