91Ó°ÊÓ

The given data is listed below as-

The photon emitted in the Balmer series is the least energetic.

Theenergy differenceis given by the equation-

$\mathbf{∆}\mathit{E}\mathbf{=}\left(13.6eV\right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Here, n is the quantum number.

The difference between energies is given by the equation:

$$\begin{gathered} ∆E=E_3-E_2 \\ ∆E=-\left(13.6eV\right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \end{gathered}$$

For, $n^2=3$and $n_1=2$

$$\begin{gathered} ∆E=-\left(13.6eV\right)\left(\frac{1}{3^2}-\frac{1}{2^2}\right) \\ =1.889eV \end{gathered}$$

Now, hc = 1240 eV.nm

$$\begin{gathered} \mathrm{λ}=\frac{hc}{∆E} \\ =\frac{1240eV.nm}{1.889eV} \\ =658nm \end{gathered}$$

Thus, the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines is 658 nm.

The difference between energies is given by the equation:

$$\begin{gathered} ∆E=E_{∞}-E_2 \\ ∆E=-\left(13.6eV\right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \end{gathered}$$

For, $n_2=∞$and $n_1=2$

$$\begin{gathered} ∆E=-\left(13.6eV\right)\left(\frac{1}{{∞}^2}-\frac{1}{2^2}\right) \\ =3.40eV \end{gathered}$$

Now, hc = 1240 eV.nm

$$\begin{gathered} \mathrm{λ}=\frac{hc}{∆E} \\ =\frac{1240eV.nm}{3.40eV} \\ =366nm \end{gathered}$$

Thus, the wavelength of the series limit is 366 nm.