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The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215Å`B. `1640Å`C. `2430Å`D. `4687Å` |
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Answer» Correct Answer - A The wavelength `(lambda)` of a spectral line in the Balmer series is given by `1/lambda=RZ^(2)[1/(2^(2))-1/(n^(2))]` where n=3,4,5,6..... In case of hydrogen atom, for first spectral line in the Balmer series, Z=1, n=3 , `lambda=lambda_(H)` (say) `:. 1/(lambda_(H))=Rxx(1)^(2)[1/(2^(2))-1/(3^(2))]=Rxx5/36.......(i)` In case of singly -ionized helium atom, for second spectral line in the Balmer series, Z=2 , n=4, `lambda=lambda_(He^(+))` `1/(lambda_(He^(+)))=Rxx(1)^(2)[1/(2^(2))-1/(4^(2))]= 4Rxx3/16.......(ii)` Dividing (i) by (ii), we get `(lambda_(He^(+)))/(lambda_(H))=(5//36)/(4xx3//16)=5/27` `lambda_(H^(+))= lambda_(H) xx5/27=(6561Å)xx5/27=1215Å` |
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