Question

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is(a) $\frac{9}{4}$(b) $\frac{27}{5}$(c) $\frac{5}{27}$(d) $\frac{4}{9}$

Answer 1

need points to ask questions so giving such answer

Answer 2

Answer:

C) 5/27

Explanation:

The wavelength of a spectral line in the Lyman series -

1/λl = R ( 1/1²-1/n²), n = 2,3,4

The wavelength of a spectral line in the Balmer series -

1/λb = R ( 1/2²-1/n²), n = 3,4, 5

For the longest wavelength in Lyman series, n = 2

1/λl = ( 1/1²-1/2²)

= R (1/1-1/4)

= R ( 4-1/4)

= 3R/4

or λl = 4/3R

For the longest wavelength in Balmer series, n = 3

1/λb = ( 1/2²-1/3²)

= R (1/4-1/9)

= R ( 9-4/36)

= 5R/36

or λb = 36/5R

Thus, λl/λb

= 4/3R × 5R/36

= 5/27

Thus the longest wavelength is 5/27.