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Calculate the wave number and wave length of H_(alpha) line in the Balmer series of hydroge emission spectrum. |
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Answer» Solution :For `H_(a)` line in the Balmer series `n_(1)=2 and n_(2)=3` `BARV=109677((1)/(2^(2))-(1)/(3^(2)))=15,232.9cm^(-1)` `lambda=(1)/(uspilon)=(1)/(15,232.9)=6.564xx10^(-5)cm` `lambda=(1)/(barv)=(1)/(15,232.9)=6.564xx10^(-5)cm` `"Wave number of "H_(ALPHA)" line" = 1.5239 XX 10^(4)cm^(-1)` `"Wave length of "H_(alpha)" line "= 6564 Å` |
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