Solution: Determining the Radius of an Excited $Be^{3+}$ Ion

This problem involves understanding the energy levels and radii of hydrogen-like ions, specifically the $Be^{3+}$ ion. We are given that the ion starts in the second excited state, absorbs a photon of a specific wavelength, and we need to find the radius of the ion in its new, higher energy state.

Understanding Atomic States

In atomic physics, the principal quantum number, denoted by '$n$', describes the energy level of an electron. The states are numbered starting from $n=1$ for the ground state (lowest energy level).

• Ground State: $n=1\)
• First Excited State: $n=2\)
• Second Excited State: $n=3\)

Therefore, the $Be^{3+}$ ion initially is in the state with $n_{initial} = 3$.

Calculating Photon Energy

The energy of the absorbed photon ($E_{photon}$) is related to its wavelength ($\lambda$) by the formula:

\[ E_{photon} = \frac{hc}{\lambda} \]

Given:

• Wavelength, $\lambda = 601.6\text{ A}$
• Planck's constant times the speed of light, $hc = 12500\text{ eV-A}$

Plugging in the values:

\[ E_{photon} = \frac{12500\text{ eV-A}}{601.6\text{ A}} \]

\[ E_{photon} \approx 20.7779\text{ eV} \]

This is the energy absorbed by the $Be^{3+}$ ion.

Calculating Energy Levels for $Be^{3+}$

Beryllium (Be) has an atomic number $Z = 4$. The $Be^{3+}$ ion has lost 3 electrons, leaving only one electron. This makes it a hydrogen-like ion, and its energy levels can be calculated using the Bohr model formula:

\[ E_n = - \frac{Z^2 \cdot 13.6\text{ eV}}{n^2} \]

where $13.6\text{ eV}$ is the ionization energy of hydrogen in its ground state.

Initial Energy State

The ion starts in the second excited state, where $n_{initial} = 3$. For $Be^{3+}$, $Z = 4$. The initial energy ($E_{initial}$) is:

\[ E_{initial} = E_3 = - \frac{4^2 \cdot 13.6\text{ eV}}{3^2} = - \frac{16 \times 13.6\text{ eV}}{9} \]

\[ E_{initial} \approx - \frac{217.6}{9} \approx -24.1778\text{ eV} \]

Final Energy State

After absorbing the photon, the ion transitions to a higher energy level. The final energy ($E_{final}$) is the sum of the initial energy and the photon energy:

\[ E_{final} = E_{initial} + E_{photon} \]

\[ E_{final} \approx -24.1778\text{ eV} + 20.7779\text{ eV} \]

\[ E_{final} \approx -3.3999\text{ eV} \]

Determining the Final Quantum Number

We need to find the principal quantum number ($n_{final}$) corresponding to this final energy level. Using the energy level formula again:

\[ E_{final} = - \frac{Z^2 \cdot 13.6\text{ eV}}{n_{final}^2} \]

\[ -3.3999\text{ eV} \approx - \frac{4^2 \cdot 13.6\text{ eV}}{n_{final}^2} = - \frac{16 \times 13.6\text{ eV}}{n_{final}^2} = - \frac{217.6\text{ eV}}{n_{final}^2} \]

Solving for $n_{final}^2$:

\[ n_{final}^2 \approx \frac{217.6\text{ eV}}{3.3999\text{ eV}} \approx 64.0016 \]

Taking the square root:

\[ n_{final} \approx \sqrt{64.0016} \approx 8 \]

So, the resulting excited state has a principal quantum number $n_{final} = 8$.

Calculating the Radius in the Resulting State

The radius ($r_n$) of the $n$-th energy level for a hydrogen-like ion with atomic number $Z$ is given by the formula:

\[ r_n = \frac{n^2}{Z} a_0 \]

where $a_0$ is the Bohr radius.

For the final state, we have $n = n_{final} = 8$ and $Z = 4$ for $Be^{3+}$.

\[ r_{final} = \frac{8^2}{4} a_0 = \frac{64}{4} a_0 \]

\[ r_{final} = 16 a_0 \]

Conclusion

The radius of the $Be^{3+}$ ion in the resulting excited state is $16 a_0$. This corresponds to option 3.