Question

calculate the shortest and longest wavelength limits os paschen series​

Answer 1

Explanation:

For Lyman series, n1=1.

For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2=∞.

So, λ1=RH[121−∞21]=RH

λ=1096781=9.117×10−6cm

       =911.7A˚

For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n2=2.

So, λ1=RH[121−221]=43RH

or λ=34×RH1=

Answer 2

Answer:

For Lyman series, n1=1.

For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2=∞.

So, λ1=RH[121−∞21]=RH

λ=1096781=9.117×10−6cm

      =911.7A˚

For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n2=2.

So, λ1=RH[121−221]=43RH

or λ=34×RH1=

Explanation: