The first member of the paschen series in hydrogen spectrum is of wavelength `18,800 Å`. The short wavelength limit of Paschen series is

Options

A. `1215 Å`
B. `6560 Å`
C. `8225 Å`
D. `12850 Å`

Answer

C

Explanation

For Paschen series `bar(v) = (1)/(lambda) = R[(1)/(3^(2)) - (1)/(n^(2))], n = 4, 5, 6..`
For first mumber of Paschen series `n = 4`
`(1)/(lambda_(1)) = R[(1)/(3^(2)) - (1)/(4^(2))] rArr (1)/(lambda_(1)) = (7R)/(144)`
`rArr R = (144)/(7lambda_(1)) = (144)/(7 xx 18800 xx 10^(-10)) = 1.1 xx 10^(-7)`
For shortest wave length `n = oo`
So `(1)/(lambda) = R [(1)/(3^(2)) - (1)/(oo^(2))] = (R )/(9)`
`rArr lambda = (9)/(R ) = (9)/(1.1 xx 10^(-7)) = 8.225 xx 10^(-7)m = 8225 Å`