Question

A triply ionized beryllium ion, Be${}^{3+}$ (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of Be${}^{3+}$? How does this compare to the ground-level energy of the hydrogen atom? (b) What is the ionization energy of Be${}^{3+}$? How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the $n$ = 2 to $n$ = 1 transition is 122 nm (see Example 39.6). What is the wavelength of the photon emitted when a Be${}^{3+}$ ion undergoes this transition? (d) For a given value of $n$, how does the radius of an orbit in Be${}^{3+}$ compare to that for hydrogen?

Short answer

(a) -217.6 eV, 16 times more negative; (b) 217.6 eV, 16 times larger; (c) 30.5 nm; (d) \(\frac{1}{4}\) hydrogen orbit.

Step-by-step solution

Energy Levels Calculation

For hydrogen-like ions, the energy of the $n$-th level is given by $-13.6\text{eV}\frac{Z^2}{n^2}$. For Be${}^{3+}$, the nuclear charge is $Z=4$. Thus, the ground level energy $E_1$ can be calculated as follows: $-13.6\text{eV}\frac{4^2}{1^2}=-13.6\text{eV}\times 16=-217.6\text{eV}$. This is 16 times more negative than the ground level energy of the hydrogen atom.

Ionization Energy

Ionization energy is the amount of energy needed to remove the electron from the ground state. For Be${}^{3+}$, the ionization energy is the same as its ground level energy (with a positive sign). Therefore, the ionization energy is $217.6\text{eV}$, which is 16 times the ionization energy of hydrogen (-13.6 eV").

Wavelength Calculation for Transition

The wavelength $\lambda$ of the photon emitted during a transition from $n=2$ to $n=1$ can be found using the formula: $s\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ where $R=1.097\times {10}^7{\text{m}}^{-1}$ and $n_1=1$, $n_2=2$. Plugging in $Z=4$ for Be${}^{3+}$: \ $\frac{1}{\lambda }=1.097\times {10}^7{\text{m}}^{-1}\times 16(\frac{1}{1^2}-\frac{1}{2^2})$ Solving it gives $\lambda =30.5\text{nm}$.

Radius Comparison

For hydrogen-like ions, the radius of an orbit is given by \ $r_n=\frac{n^2{a}_0}{Z}$where $a_0$ is the Bohr radius. For Be${}^{3+}$ with $Z=4$, the radius of an orbit is thus \ he fraction of the hydrogen orbit, $\frac{n^2{a}_0}{4}$. Thus, an orbit for Be${}^{3+}$ is times the radius in the hydrogen atom for the same $n$.

Key concepts

Beryllium ion

A Beryllium ion, specifically Be${}^{3+}$, is a fascinating species, especially when we consider its behavior like a "hydrogen-like ion." This happens when a beryllium atom loses three of its electrons, leaving it with a single electron orbiting the nucleus.

Since Beryllium has an atomic number of 4, its nucleus is four times more positively charged than that of hydrogen. This increased nuclear charge results in more tightly bound electrons compared to hydrogen, affecting various atomic properties such as energy levels and ionization potential.

• The comparison with hydrogen arises due to their similar single-electron structure in the ionized form.
• In the context of this exercise, a Be${}^{3+}$ ion offers a unique look at how increased nuclear charge affects atomic behavior compared to hydrogen.

Ground-level energy

Ground-level energy refers to the energy of an electron in its lowest energy state, or the closest permitted orbit to the nucleus. In hydrogen-like ions such as Be${}^{3+}$, this energy is calculated using the formula $-13.6\text{eV}\frac{Z^2}{n^2}$.

For Be${}^{3+}$, the charge $Z$ is 4, therefore:

$$E_1=-13.6\text{eV}\times \frac{4^2}{1^2}=-217.6\text{eV}$$

• This energy is 16 times more negative than that of hydrogen's ground level energy $-13.6\text{eV}$ because the energy depends on the square of the nuclear charge.
• The more negative value indicates a stronger attraction between the electron and the nucleus in Be${}^{3+}$.

Ionization energy

Ionization energy is the amount of energy required to remove the lone electron from its ground state orbit in an atom or ion. For hydrogen-like ions, it is equivalent to the magnitude of the ground level energy but with a positive sign.

For Be${}^{3+}$, the ionization energy is calculated as:

$$217.6\text{eV}$$

• This shows that ionizing Be${}^{3+}$ requires far more energy than ionizing hydrogen, reflecting the stronger electrostatic forces due to larger nuclear charge.
• This energy is 16 times the ionization energy of a hydrogen atom $13.6\text{eV}$.

Photon wavelength

When a photon is emitted due to an electron transition between energy levels, its wavelength can be calculated using the wavelength formula associated with transitions in hydrogen-like ions.

For a transition from $n=2$ to $n=1$, the wavelength $\lambda$ is given by:

$$\frac{1}{\lambda }=R{Z}^2(\frac{1}{1^2}-\frac{1}{2^2})$$

Plugging in the values for Be${}^{3+}$ (where $Z=4$ and $R$ is the Rydberg constant $1.097\times {10}^7{\text{m}}^{-1}$) gives a wavelength of:

$$\lambda =30.5\text{nm}$$

• This value is significantly shorter than the 122 nm for hydrogen, showing that larger nuclear charges result in higher energy transitions and shorter wavelength emissions.

Bohr radius

The Bohr radius is a crucial concept in atomic physics representing the size of the simplest orbital in the hydrogen atom. In hydrogen-like ions, it informs us about the size of the electron's orbit for the principal quantum number $n$.

For an orbit in Be${}^{3+}$, the radius $r_n$ is calculated by:

$$r_n=\frac{n^2{a}_0}{Z}$$

Here, $a_0$ is the Bohr radius for hydrogen, and $Z$ is the atomic number 4 for Be${}^{3+}$. This results in:

$$r_n=\frac{n^2{a}_0}{4}$$

• The radius is 1/4 that of the corresponding orbital in hydrogen, reflecting how increased nuclear charge contracts the electron's orbit around the nucleus.

Atomic transitions

Atomic transitions occur when electrons move between different energy levels within an atom. These transitions are responsible for the emission or absorption of light at particular wavelengths.

For Be${}^{3+}$, atomic transitions provide insights into its behavior compared to hydrogen due to its higher nuclear charge.

• The transition from $n=2$ to $n=1$ in Be${}^{3+}$ results in the emission of a photon with a 30.5 nm wavelength, which is much shorter than that of hydrogen. This points to higher energy differences between levels.
• Such transitions, accompanied by the emission of photons, are fundamental to understanding atomic emissions in spectroscopy.