Question

For the electronic transition from \(\mathrm{n}=2 \rightarrow \mathrm{n}=1\), which of the following will produce shortest wave length? (a) \(\mathrm{Li}^{2+}\) ion (b) D atom (c) \(\mathrm{He}^{+}\) ion (d) \(\mathrm{H}\) atom

Short answer

\(\mathrm{Li}^{2+}\) ion produces the shortest wavelength.

Step-by-step solution

Understand the Problem

We need to determine which of the atomic systems given (\(\mathrm{Li}^{2+}\), \(D\), \(\mathrm{He}^{+}\), \(\mathrm{H}\)) undergoing a transition from \(n=2\) to \(n=1\) produces the shortest wavelength. The shortest wavelength corresponds to the greatest change in energy between these levels.

Energy Level Formula

The energy of an electron in the \(n\)th level of a hydrogen-like atom is given by:\[ E_n = -\frac{Z^2 R_H}{n^2} \]where \(Z\) is the atomic number and \(R_H\) is the Rydberg constant for hydrogen.

Calculate Energy Levels for Each Ion/Atom

For each ion/atom:- \(\mathrm{Li}^{2+}\): \(Z = 3\); \(E_1 = -9R_H\), \(E_2 = -2.25R_H\)- \(D\) and \(\mathrm{H}\): \(Z = 1\); \(E_1 = -R_H\), \(E_2 = -0.25R_H\)- \(\mathrm{He}^{+}\): \(Z = 2\); \(E_1 = -4R_H\), \(E_2 = -R_H\)

Energy Transitions and Wavelength

The change in energy during a transition from \(n_2\) to \(n_1\) is\[ \Delta E = E_1 - E_2 \]The wavelength \(\lambda\) of the photon emitted is given by:\[ \lambda = \frac{hc}{\Delta E} \]To find the shortest wavelength, we need the largest \(\Delta E\).

Calculate Energy Changes and Compare

Calculate \(\Delta E\) for each case:- \(\mathrm{Li}^{2+}\): \(\Delta E = -9R_H + 2.25R_H = 6.75R_H\)- \(D\) and \(\mathrm{H}\): \(\Delta E = -R_H + 0.25R_H = 0.75R_H\)- \(\mathrm{He}^{+}\): \(\Delta E = -4R_H + R_H = 3R_H\)The highest \(\Delta E\) indicates the shortest wavelength.

Conclusion

\(\mathrm{Li}^{2+}\) produces the shortest wavelength for this transition as it has the largest energy change (\(6.75R_H\)).

Key concepts

Hydrogen-like Atoms

Hydrogen-like atoms are unique because they resemble the hydrogen atom in terms of their electronic structure. These atoms or ions contain only a single electron, just like hydrogen. However, their atomic nucleus may have different numbers of protons, signified as the atomic number, \(Z\). This characteristic affects their energy levels dramatically. Consider examples like the \(\mathrm{He}^{+}\) ion or \(\mathrm{Li}^{2+}\) ion. Each of these has only one electron orbiting around the nucleus but with a different number of protons compared to a hydrogen atom. This results in different energy calculations and spectrum properties. These variations are mostly due to the stronger nuclear charge in ions like \(\mathrm{He}^{+}\) or \(\mathrm{Li}^{2+}\), as compared to hydrogen with only one proton in its nucleus.Understanding hydrogen-like atoms helps in comprehending complex atomic structures by providing a simpler, more relatable reference point grounded in the well-known behavior of hydrogen.

Energy Levels

Energy levels in atoms refer to the permissible energy states that an electron can occupy around the nucleus. For hydrogen-like atoms, the energy levels are defined using the formula:\[ E_n = -\frac{Z^2 R_H}{n^2} \]where \(E_n\) is the energy at a specific level \(n\), \(Z\) stands for the atomic number, and \(R_H\) is the Rydberg constant specific to hydrogen.

• The number \(n\) is an integer called the "principal quantum number." It indicates the electron cloud's size and energy.
• As \(n\) increases, the electron resides farther from the nucleus, and these levels become closer together in terms of energy differences.

The significance of understanding energy levels lies in predicting how an electron behaves within an atom and when electronic transitions happen, which are crucial for photon emissions. Grasping these concepts lays the foundation for studying topics like spectral series and chemical bonding.

Photon Emission

Photon emission occurs when an electron transitions from a higher energy level to a lower one, releasing energy in the form of a photon. The energy change \(\Delta E\) during a transition can be calculated using the formula:\[ \Delta E = E_1 - E_2 \]where \(E_1\) and \(E_2\) are the energy levels involved. This energy directly correlates to the wavelength of the emitted photon, determined by:\[ \lambda = \frac{hc}{\Delta E} \]Here, \(\lambda\) is the wavelength, \(h\) is Planck's constant, and \(c\) is the speed of light.

• Longer wavelengths correspond to smaller energy changes, and shorter wavelengths correspond to larger energy changes.
• This principle is used in spectroscopy to determine the composition and properties of distant stars and galaxies, as each atom or ion has a unique spectral fingerprint.

Understanding photon emissions not only explains visual phenomena like atomic spectra but also helps in technological advancements such as lasers and LED lights.

Atomic Spectra

Atomic spectra emerge when electrons in an atom transition between different energy levels, manifesting as distinct lines of color when viewed through a prism or spectroscope. These spectra can either be emission spectra or absorption spectra, each revealing different atomic features.

• Emission Spectra: Produced when electrons fall to lower energy levels, emitting photons. Each line in this spectrum corresponds to a specific energy change and, consequently, a specific wavelength.
• Absorption Spectra: Occur when electrons absorb energy and move to higher energy levels. The spectrum shows dark lines at the wavelengths absorbed.

Atomic spectra act like an atomic fingerprint, allowing scientists to identify elements present in stars and galactic clouds, contribute to our understanding of the universe, and further our knowledge of quantum mechanics. These spectra stand at the heart of inventions like spectrometers and have significantly advanced our insights into chemical analysis.