1.

If the wavelength of the first line of the Balmer series of hydrogen is 6561 A, the wavelength of the second line of the series should be.

Answer»

`13122 Å`
`3280Å`
`4860Å`
`2178Å`

Solution :ln Balmer series `(1)/(lambda_(1)) = (1)/(6561Å) = R [(1)/((2)^(2)) - (1)/((3)^(2))] = (5)/(36) R""……..(i)`
and `(1)/(lambda_(2)) = R [(1)/((2)^(2)) - (1)/((4)^(2))] = (3)/(16)R""…..(II)`
`RARR ""(lambda_(2))/(6561 Å) = (5)/(36) xx (16)/(3) = (20)/(27) = or lambda_(2) = (20)/(27) xx 6561 = 4860 Å`


Discussion

No Comment Found

Related InterviewSolutions