Calculate the “series limit” of the Lyman series of spectral lines. This is defined as the shortest wavelength possible of a photon emitted in a transition from a higher initial energy level to the ${\mathbf{n}}_{\mathbf{i}}\mathbf{=}\mathbf{1}$ final level. (Note: In figure 7.5, the spectral lines of the series “crowd together” at the short-wavelength end of the series).

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, is inversely proportional to the wavelength.

The energy of the emitted photon is,

$\mathbf{E}\mathbf{=}\frac{\mathbf{hc}}{\mathbf{\lambda }}$ ….. (1)

Where, h is Plank’s constant, c is the speed of light, and role="math" localid="1659178129240" $\mathbf{\lambda }$is the wavelength of the photon.

The Planck’s constant, $\mathrm{h}=1240\mathrm{eV}·\mathrm{nm}$

From eq. (1) you get that energy is high if the wavelength is low.

Hence, shortest wavelength will be because of the highest energy possible in the Lyman series, i.e. $\mathrm{n}=\infty \to \mathrm{n}=1$,

Where, n is the principal quantum number.

As you know that energy of a state is

${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}(\mathrm{n}=12,3,\dots \dots )$

Where, n is the principal quantum number.

Hence, Energy corresponding to $\mathrm{n}=\infty \to \mathrm{n}=1$will be given by,

$\begin{array}{l}\mathrm{E}={\mathrm{E}}_{\infty }-{\mathrm{E}}_1\\ =-\frac{13.6\mathrm{eV}}{{\left(\infty \right)}^2}+\frac{13.6\mathrm{eV}}{{\left(1\right)}^2}\\ =0+\frac{13.6\mathrm{eV}}{1}\end{array}$

$\mathrm{E}=1.36\mathrm{eV}$ ….. (2)

Now, substitute $1240\mathrm{eV}·\mathrm{nm}$for localid="1659179281565" $\mathrm{h}$and $\mathrm{E}=1.36\mathrm{eV}$for E into equation (1).

$$\begin{gathered} 1.36\mathrm{eV}=\frac{1240\mathrm{eV}·\mathrm{nm}}{\mathrm{\lambda }} \\ \mathrm{\lambda }=91.2\mathrm{nm} \end{gathered}$$

Hence, required shortest wavelength of the Lyman series is 91.2 nm.