Question

when an electron de-excites from higher orbit in H-atom, only two radiations are emitted out one in Paschen series and one in Lyman series. The wavelength of radiation emitted out in Lyman series is?

• A. 8R/9
• B. 3R/4
• C. 4/3R
• D. 9/8R

Answer 1

Answer: (D)

9/8R

Answer 2

The electron de-excites in a cascade: $n_{initial} \rightarrow 3$ (Paschen series) and then $3 \rightarrow 1$ (Lyman series). The wavelength of the Lyman series radiation corresponds to the $3 \rightarrow 1$ transition. Using the Rydberg formula $\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2})$ for $n_f=1$ and $n_i=3$, we get $\frac{1}{\lambda} = R(\frac{1}{1^2} - \frac{1}{3^2}) = R(1 - \frac{1}{9}) = R(\frac{8}{9})$. Thus, $\lambda = \frac{9}{8R}$.