The first line of Balmer series has wavelength `6563 Å`. What will be the wavelength of the first member of Lyman series?

Options

A. `1215 Å`
B. `4861 Å`
C. `4340 Å`
D. `4101 Å`

Answer

A

Explanation

For first line Balmer series.
`(1)/(lambda) = R[(1)/(2^(2)) -(1)/(3^(2))]`
or `(1)/(lambda) = R[(1)/(4) -(1)/(9)]`
or `(1)/(lambda) = R[(9-4)/(36)] = (5R)/(36) lambda = (36)/(5R)`
for first line Lyman series `(1)/(lambda') = R [(1)/(1^(2)) -(1)/(2^(2))]`
`(1)/(lambda') = R[1-(1)/(4)]`
`(1)/(lambda') = (3R)/(4)` or `lambda' = (4)/(3R)`
or `(lambda')/(lambda) = (4)/(3R) xx (5R)/(36) = (20)/(108)`
`lambda' = (20)/(108) xx 6563 Å`
`= 1215 Å`