Question

1. What are the initial and final energy levels for the third (i.e., third-longest wavelength) line in the Paschen series? (See Figure 7.5)
2. Determine the wavelength of this line.

Short answer

(a) Energy of initial state is $-0.378\mathrm{eV}$and Energy of final state $-1.512\mathrm{eV}$.

(b) Wavelength of the third line of the Paschen series is $1.1x{10}^{-6}$m.

Step-by-step solution

The Paschen series:

The Paschen series is a set of emission lines from atomic hydrogen gas due to electrons descending from an electron shell with n greater than 3 up to n = 3, or analogous absorption lines when absorbed electromagnetic radiation causes the electrons to do the opposite.

(a) Initial and final energy levels of the third line in the Paschen series:

All the spectrum lines which originate on n = 4,5,6,7,... and end at n = 3 are said to be in the Paschen series of Spectral lines. Where, n is the principal quantum number.

As you know, the third line of the Paschen series has an initial state, n = 6, and the final state as n = 3.

You also know that Energies of different quantized levels of the hydrogen atom can be given by

_{${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6\mathrm{eV}}{{\left(3\right)}^2}\left(n=1,2,3,....\right)$}

Now, energy of initial state (n = 6) :

$$\begin{gathered} E_6=-\frac{13.6\mathrm{eV}}{{\left(6\right)}^2} \\ =-0.378\mathrm{eV} \end{gathered}$$

Now, energy of final state ( n = 3) :

_{$$\begin{gathered} E_6=-\frac{13.6\mathrm{eV}}{{\left(3\right)}^2} \\ =-1.512\mathrm{eV} \end{gathered}$$}

(b) Wavelength of 6→3  transition:

As you know that, energy of photon emitted during $6\to 3$transition is given by,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{Photon}}={\mathrm{E}}_6-{\mathrm{E}}_3 \\ =-0.378\mathrm{eV}-\left(-1.512\mathrm{eV}\right) \end{gathered}$$

${\mathrm{E}}_{\mathrm{Photon}}=1.13\mathrm{eV}$ ….. (1)

You also know that, energy of emitted photon is given by

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

Where, h is Plank’s constant, c is the speed of light, $\lambda$is the wavelength of the photon.

Now, by using equation (1) and equation (2), you get,

$$\begin{gathered} 1.13\mathrm{eV}=\frac{1240\mathrm{eV}\mathrm{nm}}{\mathrm{\lambda }} \\ \mathrm{\lambda }=1.1\times {10}^{-6}\mathrm{m} \end{gathered}$$

Hence, wavelength of the required third line of the Paschen series will be $\mathrm{\lambda }=1.1\times {10}^{-6}\mathrm{m}$.