Question

Calculate the shortest wavelength of H-alpha line​

Answer 1

To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line. Solution: From the previous problem we know that the shortest-wavelength Balmer line has λ = 365 nm. The transition from ni = 2 to nf = 1 is the lowest energy, longest wavelength transition in the Lyman series.

Answer 2

Answer:

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Explanation:

For Lyman series n1=1n1=1

For shortest λλ of Lyman series; energy difference in two levels showing transition should be maximum i.e n2=∞n2=∞

1λ=RH[112−1∞2]1λ=RH[112−1∞2]

1λ=1096781λ=109678

λ=911.7×10−8cmλ=911.7×10−8cm

=911.7A∘=911.7A∘

For longest λλ of Lyman series; energy difference in two levels showing transition should be minimum i.e n2=2n2=2

1λ=RH[112−122]1λ=RH[112−122]

=109678×34=109678×34

λ=1215.67×10−8cmλ=1215.67×10−8cm

=1215.67A∘=1215.67A∘

Hence answer is (a)