Chapter 6, Problem 105AP

Interpretation Introduction

Interpretation:

For the first four transitions in Balmer series, the wavelength is to be calculated for ${\text{He}}^{+}$ ion. Also, the wavelengths of the same transition for Hatom are to be calculatedand a comment is to be made on the differences of the calculated wavelengths of ${\text{He}}^{+}$ ion and Hatom.

Concept introduction:

The Balmer series corresponds to transitions $n=2$.

The wavelength of the electron when it jumps from one energy state to another is determined as

$\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Here, $n_f$ is the final state, and $n_i$ is the initial state, $\lambda$ is wavelength, and $R_{\infty }$ is Rydberg constant $\left(1.0968\times {10}^7 {\text{m}}^{-1}\right)$.

The relationship between meters and nanometers can be expresses as

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

Answer to Problem 105AP

Solution: For the first four transitions in the Balmer series, the wavelength of He⁺ ion in the increasing order is $103\text{ nm}$, $108\text{ nm}$, $121\text{ nm}$, and $164\text{ nm}$.

Since the Rydberg constant for Hatom and He⁺ ion is different, Balmer transitions for He⁺ ion are in the ultraviolet region, whereas the transitions for Hatom are all in the visible region.

Explanation of Solution

Given information: $R_{\infty }\text{ for He}=4.39\times {10}^7{\text{ m}}^{-1}$.

The first four transitions in the Balmer series are as follows:

$\begin{array}{r}{1}^{\text{st}}\text{ transition: }{n}_i=3\text{ to }{n}_f=2\\ 2^{\text{nd}}\text{ transition: }{n}_i=4\text{ to }{n}_f=2\\ 3^{\text{rd}}\text{ transition: }{n}_i=5\text{ to }{n}_f=2\\ 4^{\text{th}}\text{ transition: }{n}_i=6\text{ to }{n}_f=2\end{array}$

For the firsttransition $\left(n_i^{}=3\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=4.39\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ \lambda =1.64\times {10}^{-7}\text{ m }\\ \lambda =164\text{ nm}\end{array}$

For the second transition $\left(n_i^{}=4\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=4.39\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\\ \lambda =1.21\times {10}^{-7}\text{ m}\\ \lambda =121\text{ nm}\end{array}$

For the third transition $\left(n_i^{}=5\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=4.39\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\\ \lambda =1.08\times {10}^{-7}\text{ m }\\ \lambda =108\text{ nm}\end{array}$

For the fourth transition $\left(n_i^{}=6\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=4.39\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{6^2}\right)\\ \lambda =1.03\times {10}^{-7}\text{ m}\\ \lambda =103\text{ nm}\end{array}$

Therefore, for the first four transitions in the Balmer series, the wavelength of He⁺ ion in the increasing order is $103\text{ nm}$, $108\text{ nm}$, $121\text{ nm}$, and $164\text{ nm}$.

The Rydberg constant for Hatom is $1.0968\times {10}^7 {\text{m}}^{-1}$. Now, the wavelength of Hatom for the first transition of the Balmer series is as

For the first transition $\left(n_i^{}=3\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=1.0968\times {10}^7 {\text{m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ \lambda =6.56\times {10}^{-7}\text{ m}\\ \lambda =656\text{ nm}\end{array}$

For the second transition $\left(n_i^{}=4\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=1.0968\times {10}^7 {\text{m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\\ \lambda =4.86\times {10}^{-7}\text{ m}\\ \lambda =486\text{ nm}\end{array}$

For the third transition $\left(n_i^{}=5\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=1.0968\times {10}^7 {\text{m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\\ \lambda =4.34\times {10}^{-7}\text{ m}\\ \lambda =434\text{ nm}\end{array}$

For the fourth transition $\left(n_i^{}=6\text{ to }{n}_f^{}=2\right)$,

$\begin{array}{c}\frac{1}{\lambda }=R_{\infty }\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ \frac{1}{\lambda }=1.0968\times {10}^7 {\text{m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{6^2}\right)\\ \lambda =4.10\times {10}^{-7}\text{ m}\\ \lambda =410\text{ nm}\end{array}$

Therefore, for the first four transitions in the Balmer series, the wavelength of Hatom in the decreasing order is $656\text{ nm}$, $486\text{ nm}$, $434\text{ nm}$, and $410\text{ nm}$.

There is a huge difference between the wavelengths of Hatom and He⁺ ion for the first four transitions of the Balmer series, because the Rydberg constant for both of them is different, and all the Balmer transitions for He⁺ ion are in the ultraviolet region, whereas the transitions for Hatom are all in the visible region.

Conclusion

Since the Rydberg constant for Hatom and He⁺ ion is different, there is a huge difference between the wavelengths of Hatom and He⁺ ion for the first four transitions of the Balmer series. Balmer transitions for He⁺ ion are in the ultraviolet region, whereas the transitions for Hatom are all in the visible region.