1.

The `lambda` of `H_(alpha)` line of the Balmer series is `6500 Å` What is the `lambda` of `H_(beta)` line of the Balmer series

Answer» For `H_(alpha)`line of the balmer series `n_(1) = 2,n_(2) = 3`
For `H_(beta)`line of the balmer series `n_(1) = 2,n_(2) = 4`
`1/(lambda_(H_(alpha))) = R_(H) [(1)/(2^(2)) - (1)/(3^(2))]`……..(i)
` and `1/(lambda_(H_(beta))) = R_(H) [(1)/(2^(2)) - (1)/(4^(2))]`……..(ii)
By equation (i) and (ii) we get
`:. (lambda_(beta))/(lambda_(alpha))= ((1)/(4) - (1)/(9))/((1)/(4) - (1)/(16))`
`:. lambda_(beta) = lambda_(alpha) xx [(80)/(108)] = 6500 xx (80)/(108) = 4814.8 Å`


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