Question

If λ_max for Lyman series of H-atom is 912 Å, then calculate λ_min for Balmer series of H-atom (in Å). 

Answer 1

Correct answer is:  2736

λ_max for Lyman series (E = 2 →  E = 1)

\(\frac{1}{912}=R(1)^2\left(\frac{1}{4}-\frac{1}{4}\right)\)

\(\frac{1}{912}=R\times\frac{3}{4}\)

R = \(\frac{4}{912\times3}\)

λ_min for Balmer series (E= ∞ → E = 1)

\(\frac{1}{λ}=R(1)\left(\frac{1}{4}\right)\)

\(= \frac{4}{912\times3}\times\frac{1}{4}\)

\(= \frac{1}{912\times3}\)

\(λ=912\times3\)

= 2736 Å