Question

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

Answer 1

As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers n = 2 to n = 1.
Wavelength of radiation $\left(\lambda \right)$ is given by
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here, R is the Rydberg constant, having the value of 1.097×10⁷ m^-1.

$$\begin{gathered} \frac{1}{\lambda }=1.097\times {10}^7\left[\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right] \\ \Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left[1-\frac{1}{4}\right] \\ \Rightarrow \frac{1}{\lambda }=1.097\times \frac{3}{4}\times {10}^7 \\ \Rightarrow \lambda =\frac{4}{1.097\times 3\times {10}^7} \\ =1.215\times {10}^{-7} \\ =121.5\times {10}^{-9}=122\mathrm{nm} \end{gathered}$$