Chapter 6, Problem 98AP

Interpretation Introduction

Interpretation:

The fact that the spectral lines of Lyman and Balmer series do not overlap each other is to be verified by calculating longest wavelength associated with Lyman series and shortest wavelength associated with Balmer series.

Concept introduction:

Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron go from $n\ge 2ton=1$(where n is the principal quantum number), the lowest energy level of the electron.

On the other hand, the series of spectral emission lines of the hydrogen atom that result from electron transitions from higher levels down to the energy level with principal quantum number $2$ is known as Balmer series.

The energy of the electron in hydrogen atom is as follows:

$E_n=-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{n^2}\right)$

Here, n is considered as an integer and $E_n$ is the energy of the electron.

The energy difference of the two states of an element is represented as:

$\Delta E=hc\lambda$

Here, h is the Planck’s constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Answer to Problem 98AP

Solution: The spectral lines of Lyman and Balmer series do not overlap each other.

Explanation of Solution

The Planck’s constant is $6.6\times {10}^{-34}\text{ J}\text{.s}$.

The speed of light is ${\text{3×10}}^{\text{8}}\text{ m}/\text{s}$.

The Lyman series deals with the excitation in first orbit.

The energy for the longest wavelength will be when the electron jumps from the second orbit to the Lyman series orbit.

So, the initial state $n_{\text{i}}=2$ and final state $n_{\text{f}}=1$.

The energy of the electron in Lyman series orbit will be as follows:

$\Delta E=-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^{\text{2}}}\right)$

Substitute 2 for $n_{\text{i}}$ and $1$ for $n_{\text{f}}$ in the above equation

$\begin{array}{c}\Delta E=-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ =-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{1}-\frac{1}{4}\right)\\ =-2.18\times {10}^{-18}\text{ J }\left(-\frac{1}{12}\right)\\ =-1.64\times {10}^{-18}\text{ J }\end{array}$

Calculate the wavelength as follows:

$\Delta E=hc\lambda$

Rearrange the above equation as follows:

$\lambda =\frac{hc}{\Delta E}$

Substitute $6.6\times {10}^{-34}\text{ J}\text{.s}$ for $h$, ${\text{3×10}}^{\text{8}}\text{ m}/\text{s}$ for $c$, and $-1.64\times {10}^{-18}\text{ J }$ for $\Delta E$ in the above equation

$\begin{array}{c}\lambda =-\frac{\left(6.6\times {10}^{-34}\text{ J}\text{.s}\right)\left({\text{3×10}}^{\text{8}}\text{ m}/\text{s}\right)}{1.64\times {10}^{-18}\text{ J }}\\ =1.21\times {10}^{-7}\text{ m}\end{array}$

Convert meter to nanometer as follows:

$\begin{array}{c}1.21\times {10}^{-7}\text{ m}=\left(1.21\times {10}^{-7}\cancel{\text{m}}\right)\left(\frac{{\text{10}}^{-9}\text{nm}}{1\cancel{\text{m}}}\right)\\ =121\text{ nm}\end{array}$

So, the longest wavelength of Lyman series is $121\text{ nm}$.

The Balmer series deals with the excitation in second orbit.

The energy for the shortest wavelength will be when the electron jumps from the second orbit to the Lyman series orbit.

So, the initial state $n_{\text{i}}=\infty$ and final state $n_{\text{f}}=2$.

The energy of the electron in Balmer series orbit will be:

$\Delta E=-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{n_{\text{f}}^{\text{2}}}-\frac{1}{n_{\text{i}}^{\text{2}}}\right)$

Substitute $\infty$ for $n_{\text{i}}$ and $2$ for $n_{\text{f}}$ in the above equation

$\begin{array}{c}\Delta E=-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)\\ =-2.18\times {10}^{-18}\text{ J }\left(\frac{1}{4}-\frac{1}{\infty }\right)\\ =-5.45\times {10}^{-19}\text{ J }\end{array}$

Calculate the wavelength as follows:

$\Delta E=hc\lambda$

Rearrange the above equation as follows:

$\lambda =\frac{hc}{\Delta E}$

Substitute $6.6\times {10}^{-34}\text{ J}\text{.s}$ for $h$, ${\text{3×10}}^{\text{8}}\text{ m}/\text{s}$ for $c$, and $-5.45\times {10}^{-19}\text{ J }$ for $\Delta E$ in the above equation

$\begin{array}{c}\lambda =-\frac{\left(6.6\times {10}^{-34}\text{ J}\text{.s}\right)\left({\text{3×10}}^{\text{8}}\text{ m}/\text{s}\right)}{5.45\times {10}^{-19}\text{ J }}\\ =\frac{19.8\times {10}^{-26}\text{ J}.\text{m}}{5.45\times {10}^{-19}\text{ J }}\\ =3.65\times {10}^{-7}\text{ m}\end{array}$

$1\text{ m}={10}^9\text{ nm}$

Convert meter to nanometer as follows:

$\begin{array}{c}3.65\times {10}^{-7}\text{ m}=\left(3.65\times {10}^{-7}\cancel{\text{m}}\right)\left(\frac{{\text{10}}^{-9}\text{nm}}{1\cancel{\text{m}}}\right)\\ =365\text{ nm}\end{array}$

So, the shortest wavelength of Balmer series is $365\text{ nm}$.

Conclusion

The spectral lines of Lyman and Balmer series do not overlap each other.