Chapter 7, Problem 7.112QP

The He⁺ ion contains only one electron and is therefore a hydrogenlike ion. Calculate the wavelengths, in increasing order, of the first four transitions in the Balmer series of the He⁺ ion. Compare these wavelengths with the same transitions in a H atom. Comment on the differences. (The Rydberg constant for He⁺ is 8.72 × 10⁻¹⁸ J.)

Interpretation Introduction

Interpretation:

The wavelengths in the increasing order of the first four transitions in the Balmer series of the ${\text{He}}^{\text{+}}$ ion, comparing these wavelengths with the same transitions in an $\text{H}$ atom and the comment on the differences between these wavelengths of ${\text{He}}^{\text{+}}$ ion and $\text{H}$ atom which occur in the electromagnetic radiation should be described by applying the Rydberg equation in them.

Concept Introduction:

Absorption refers to how much light can be taken in by the material being measured.

Emission spectrum:

When electromagnetic radiation interacts with matter, atoms and molecules may absorb energy and reach to a higher energy state.  With higher energy, these are in an unstable state.  For returning to their normal (more stable, lower energy) energy state, the atoms and molecules emit radiations in various regions of the electromagnetic spectrum.  The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum.

In 1885, Johann Balmer developed a simple equation which could be used to calculate the wavelengths of the four visible lines in the emission spectrum of hydrogen.  Johannes Rydberg developed Balmer’s equation further, giving an equation which could calculate the visible wavelengths and also those of all hydrogen’s spectral lines.

$\frac{\text{1}}{\text{λ}}={\text{ R}}_{\infty }\left(\frac{\text{1}}{{\text{n}}_{\text{1}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}{}^{\text{2}}}\right)$

This equation is known as the Rydberg equation.  Here, $\text{λ}$ is the wavelength of a line in the spectrum; ${\text{R}}_{\infty }$ is the Rydberg constant ($\text{1}{\text{.09737316 × 10}}^{\text{7}}{\text{ m}}^{-1}$ for $\text{H}$ atom); and ${\text{n}}_{\text{1}}{\text{ and n}}_{\text{2}}$ are positive integers, where ${\text{n}}_{\text{2}}{\text{ > n}}_{\text{1}}$.

Answer to Problem 7.112QP

The wavelengths of the first four transitions in the Balmer series of the ${\text{He}}^{\text{+}}$ ion are 164 nm, 121 nm, 108 nm and 103 nm whereas for $\text{H}$ atom, they are 656 nm, 486 nm, 434 nm and 410 nm.  All the Balmer transitions for ${\text{He}}^{\text{+}}$ are in the ultraviolet region whereas the transitions for $\text{H}$ atom are all in the visible region.

Explanation of Solution

When one of helium’s electrons is removed, the resulting species is the helium ion, ${\text{He}}^{\text{+}}$.  The ${\text{He}}^{\text{+}}$ ion contains only one electron and is therefore a “hydrogen-like ion”.  Balmer series corresponds to transitions to $\text{n }=\text{ 2}$ level.  The first four transitions in the Balmer series mean the transitions from $\text{n }=\text{ 3, 4, 5 and 6}$ energy levels to the transition $\text{n }=\text{ 2}$ energy level.  The formula to calculate the wavelength from Rydberg equation is

$\text{λ }=\frac{\text{1}}{{\text{R}}_{\infty }\left(\frac{\text{1}}{{\text{n}}_{\text{1}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}{}^{\text{2}}}\right)}$

Here, the Rydberg constant for $\text{He}$ is $\text{4}{\text{.39 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}$.

For the transition $\text{n }=\text{ 3 }\to \text{ n }=\text{ 2}$, the wavelength of the ${\text{He}}^{\text{+}}$ ion is calculated as follows:

$\begin{array}{l}\text{λ }=\frac{\text{1}}{{\text{R}}_{\infty }\left(\frac{\text{1}}{{\text{n}}_{\text{1}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}{}^{\text{2}}}\right)}\\ \text{λ }=\frac{\text{1}}{\text{4}{\text{.39 × 10}}^{\text{7}}{\text{m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{3}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=-\text{1}{\text{.64 × 10}}^{-\text{7}}\text{ m}\end{array}$

The negative sign indicates that the emission of light occurs.  Wavelengths are always positive signs.  Here, $\text{1 nm }={\text{ 10}}^{-\text{9}}\text{ m}$

$\begin{array}{l}\text{λ }=\text{ 1}{\text{.64 × 10}}^{-\text{7 }}\text{m}\\ \text{λ }=\text{ 1}{\text{.64 × 10}}^{-\text{7 }}\text{m × }\frac{{\text{10}}^{\text{9 }}\text{nm}}{\text{1 m}}\\ \text{λ }=\text{ 164 nm}\end{array}$

For the transition $\text{n }=\text{ 3 }\to \text{ n }=\text{ 2}$, the wavelength of the ${\text{He}}^{\text{+}}$ ion is 164 nm.  Simiarly, for the transition $\text{n }=\text{ 4 }\to \text{ n }=\text{ 2}$, the wavelength of the ${\text{He}}^{\text{+}}$ ion is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{4}{\text{.39 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{4}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 121 nm}\end{array}$

For the transition $\text{n }=\text{ 5 }\to \text{ n }=\text{ 2}$, the wavelength of the ${\text{He}}^{\text{+}}$ ion is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{4}{\text{.39 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{5}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 108 nm}\end{array}$

For the transition $\text{n }=\text{ 6 }\to \text{ n }=\text{ 2}$, the wavelength of the ${\text{He}}^{\text{+}}$ ion is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{4}{\text{.39 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{6}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 103 nm}\end{array}$

The Rydberg constant for $\text{H}$ atom is $\text{1}{\text{.097 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}$.  For the transition $\text{n }=\text{ 3 }\to \text{ n }=\text{ 2}$, the wavelength of the $\text{H}$ atom is calculated as follows:

$\begin{array}{l}\text{λ }=\frac{\text{1}}{{\text{R}}_{\infty }\left(\frac{\text{1}}{{\text{n}}_{\text{1}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}{}^{\text{2}}}\right)}\\ \text{λ }=\frac{\text{1}}{\text{1}{\text{.097 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{3}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=-\text{6}{\text{.56 × 10}}^{-\text{7}}\text{ m}\end{array}$

The negative sign indicates that the emission of light occurs.  Wavelengths are always positive signs.  Here, $\text{1 nm }={\text{ 10}}^{-\text{9}}\text{ m}$

$\begin{array}{l}\text{λ }=\text{ 6}{\text{.56 × 10}}^{-\text{7}}\text{ m}\\ \text{λ }=\text{ 6}{\text{.56 × 10}}^{-\text{7}}\text{m × }\frac{{\text{10}}^{\text{9}}\text{ nm}}{\text{1 m}}\\ \text{λ }=\text{ 656 nm}\end{array}$

For the transition $\text{n }=\text{ 3 }\to \text{ n }=\text{ 2}$, the wavelength of the $\text{H}$ atom is 656 nm.  Simiarly, for the transition $\text{n }=\text{ 4 }\to \text{ n }=\text{ 2}$, the wavelength of the $\text{H}$ atom is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{1}{\text{.097 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{4}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 486 nm}\end{array}$

For the transition $\text{n }=\text{ 5 }\to \text{ n }=\text{ 2}$, the wavelength of the $\text{H}$ atom is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{1}{\text{.097 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{5}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 434 nm}\end{array}$

For the transition $\text{n }=\text{ 6 }\to \text{ n }=\text{ 2}$, the wavelength of the $\text{H}$ atom is

$\begin{array}{l}\text{λ }=\frac{\text{1}}{\text{1}{\text{.097 × 10}}^{\text{7}}{\text{ m}}^{-\text{1}}\left(\frac{\text{1}}{{\text{6}}^{\text{2}}}-\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)}\\ \text{λ }=\text{ 434 nm}\end{array}$

All the Balmer transitions for ${\text{He}}^{\text{+}}$ are in the ultraviolet region whereas the transitions for $\text{H}$ atom are all in the visible region.