Problem 95 - A Complete Resource Book in Chemistry for JEE Main

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

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

The exercise asks which electronic transition from n=2 to n=1 produces the shortest wavelength. We need to evaluate this for different hydrogen-like ions and atoms.

Step 2: Identify the Formula

To find the wavelength of a transition, we use the Rydberg formula: \( \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( R \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1, n_2 \) are the principal quantum numbers of the initial and final states.

Step 3: Determine Atomic Numbers

Identify the atomic numbers of each species: \( Z = 3 \) for \( \mathrm{Li}^{2+} \), \( Z = 1 \) for D (deuterium) and \( \mathrm{H} \), and \( Z = 2 \) for \( \mathrm{He}^{+} \).

Step 4: Calculate Energy Difference

The energy difference \( \Delta E \) for the transition is given by \( RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3RZ^2}{4} \). A larger \( Z \) results in a larger energy difference, hence a shorter wavelength.

Step 5: Evaluate Each Atomic Species

Calculate \( \Delta E \) and compare: For \( \mathrm{Li}^{2+} \), \( Z=3 \), so \( \Delta E_{Li} = \frac{27R}{4} \). For \( \mathrm{He}^{+} \), \( Z=2 \), so \( \Delta E_{He} = \frac{12R}{4} \). For D and \( \mathrm{H} \), \( Z=1 \), so \( \Delta E_{H/D} = \frac{3R}{4} \).

Step 6: Conclusion

The ion with the largest \( Z \) value, \( \mathrm{Li}^{2+} \), will have the highest \( \Delta E \) and hence the shortest wavelength for the transition from \( n=2 \) to \( n=1 \).

Key Concepts

Rydberg Formula

The Rydberg Formula is a powerful equation in atomic physics, used to calculate the wavelengths of spectral lines in many chemical elements. It's particularly useful for computing electronic transitions between energy levels in hydrogen-like ions. The formula is given by:\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where:

• \( \lambda \) is the wavelength of the emitted or absorbed light.
• \( R \) is the Rydberg constant, a fundamental constant with a value of approximately \( 1.097 \times 10^7 \) per meter.
• \( Z \) is the atomic number of the species involved.
• \( n_1 \) and \( n_2 \) are the principal quantum numbers of the energy levels involved.

Understanding the Rydberg formula helps in predicting which transitions produce certain wavelengths of light, which is crucial in fields like spectroscopy.

Wavelength Calculation

Wavelength calculation in atomic transitions involves determining the wavelength of light emitted as an electron transitions between different energy levels. Using the Rydberg formula, we calculate the wavelength \( \lambda \) of light based on specific electron transitions.The formula shows that the wavelength is inversely proportional to the difference in the squares of the reciprocals of the energy levels \( n_1 \) and \( n_2 \). Therefore, larger differences between these levels result in shorter wavelengths:- When \( n_1 \) is significantly smaller than \( n_2 \), the value of \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \) increases, reducing \( \lambda \).Applying this concept helps in understanding why certain species produce shorter wavelengths, specifically in the case of ions or atoms with higher atomic numbers undergoing identical transitions.

Atomic Number Impact

The atomic number, \( Z \), plays a crucial role in the transitions of electrons within an atom. In the Rydberg formula, \( Z^2 \) is multiplied by other factors, significantly impacting the outcome.Higher atomic numbers increase the energy difference between levels, which results in shorter wavelengths:

• The term \( Z^2 \) means that even a slight increase in \( Z \) greatly enhances the energy change \( \Delta E \), making light emission more energetic and the wavelength decreasing.
• This is why ions like \( \mathrm{Li}^{2+} \) and \( \mathrm{He}^{+} \) with higher atomic numbers than hydrogen produce shorter wavelengths during transitions.

Understanding this helps predict which species will have the shortest wavelength for any given transition, confirming predictions with real-world measurements.

Energy Levels in Atoms

Energy levels in atoms are like distinct steps or orbits that an electron can occupy. The principal quantum numbers \( n \) represent these levels, starting from the innermost orbit and moving outward.- When electrons move between these levels, energy is absorbed or emitted in the form of light.- The farther the jump, either to a higher or a lower level, the more energy exchanged.For example, a transition from \( n=2 \) to \( n=1 \) releases a significant amount of energy and hence a shorter wavelength is produced. The specific levels involved in transitions dictate the characteristics of the emission or absorption spectrum of an atom. This makes energy levels in atoms fundamental to understanding atomic spectra and the nature of electronic transitions.