1.

The seond line of Balmer series has wavelength `4861 Å` The wavelength o fthe first line Balmer series isA. `1216 Å`B. `6563 Å`C. `4340 Å`D. `4101 Å`

Answer» Correct Answer - B
For first member,
`(1)/(lambda_(1)) = R[(1)/(2^(2))-(1)/(3^(2))]`
or `(1)/(lambda_(1)) = R[(1)/(4)-(1)/(9)]`
or `(1)/(lambda) = R [(9-4)/(36)]`
or `lambda_(1) = (36)/(5R)`
For second mumber,
`(1)/(lambda) = R[(1)/(2^(2))-(1)/(4^(2))]` or `(1)/(lambda_(2)) = R[(1)/(4)-(1)/(16)]`
or `(1)/(lambda_(2)) = R[(4-1)/(16)]` or `(1)/(lambda_(2)) = (3R)/(16)`
or `lambda_(2) = (16)/(3R)`
Now, `(lambda_(1))/(lambda_(2)) = (36)/(5R) xx (3R)/(16)` or `(lambda_(1))/(lambda_(2)) = (27)/(20)`
or `lambda_(1) = (27)/(20) xx lambda_(2) = (27)/(20)xx 4861 Å = 6562.4 Å` .


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