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) He' ion (d) \(\mathrm{H}\) atom

Step-by-Step Solution

Answer

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

Step 1: Understand the Concept

The energy difference between two energy levels in an atom is inversely proportional to the wavelength of the emitted or absorbed light. The greater the energy difference, the shorter the wavelength.

Step 2: Calculate Energy Transition Formula

The energy difference between two levels in a hydrogen-like atom can be expressed as: \\[ \Delta E = 13.6 \, \text{eV} \times Z^2 \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] \where \(n_1\) and \(n_2\) are the principal quantum numbers, and \(Z\) is the atomic number.

Step 3: Apply the Formula to Each Option

Describe each option: \- (a) \(\mathrm{Li}^{2+}\) has \(Z = 3\).\- (b) D atom is similar to a hydrogen atom so \(Z = 1\).\- (c) \mathrm{He}^+ has \(Z = 2\).\- (d) \mathrm{H} atom has \(Z = 1\). \Substitute \(Z\) into the formula for each ion and calculate \(\Delta E\).

Step 4: Calculate Energy Differences

Compute the energy for each electron transition using the formula from Step 2.\- For \(\mathrm{Li}^{2+}\): \[ \Delta E = 13.6 \, \text{eV} \times 3^2 \times \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \, \text{eV} \times 9 \times 0.75 = 91.8 \, \text{eV} \] \- For the \(\mathrm{D} ext{ atom}\) and \(\mathrm{H} ext{ atom}\):\[ \Delta E = 13.6 \, \text{eV} \times 1^2 \times 0.75 = 10.2 \, \text{eV} \]\- For \(\mathrm{He}^{+}\): \[ \Delta E = 13.6 \, \text{eV} \times 2^2 \times 0.75 = 54.4 \, \text{eV} \]

Step 5: Determine the Shortest Wavelength

Given that wavelength is inversely proportional to the energy change, the ion with the highest energy change produces the shortest wavelength. The highest \(\Delta E\) is for \(\mathrm{Li}^{2+}\) at 91.8 eV.

Key Concepts

Energy Levels

Atoms contain electrons that exist in various energy levels or shells around the nucleus. These energy levels are quantized, meaning electrons can only occupy specific levels and cannot exist between them.
When an electron jumps from a higher energy level to a lower one, it emits energy in the form of light.

• The energy difference between these levels determines the characteristics of the light emitted, such as its wavelength.
• The larger the difference in energy between the two levels, the more energetic and shorter the wavelength of the emitted light.

For hydrogen-like atoms, this concept of energy levels is central to understanding electronic transitions and the resulting wavelengths of emitted light.

Wavelength

Wavelength refers to the distance between two consecutive peaks of a wave. When electrons transition between different energy levels in an atom, they emit or absorb light with a specific wavelength.

• This wavelength is inversely proportional to the energy difference between the two energy levels. Hence, a larger energy difference results in a shorter wavelength.
• The relationship is described using Planck's equation, where energy and wavelength are related by a constant: \[ E = \frac{hc}{\lambda} \]

In this formula, \( E \) is energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
This understanding helps in predicting the spectral lines observed in atomic spectra.

Hydrogen-like Atoms

Hydrogen-like atoms are those that have only one electron, like hydrogen itself. This includes ions such as He extsuperscript{+} or Li extsuperscript{2+} that have been ionized to remove all but one electron.
These atoms are simpler to analyze because:

• They exhibit quantized energy levels similar to hydrogen.
• Their single electron experiences a nucleus with a charge that can be described by an effective atomic number, \( Z \), which stands for the total number of protons in the nucleus.

This makes them ideal systems for studying electronic transitions by calculating energy differences using hydrogen-like formulas.
Despite their simplicity, the patterns observed in these atoms reveal fundamental insights into atomic structure and behavior.

Principal Quantum Numbers

Principal quantum numbers, denoted as \( n \), describe the main energy levels in an atom. They are integers (1, 2, 3,...) and dictate the size and energy of an electron's orbit.
Several important points about principal quantum numbers include:

• Higher values of \( n \) correspond to energy levels that are further away from the nucleus, and thus have higher energy.
• In hydrogen-like atoms, the energy associated with a level \( n \) is given by the formula: \[ E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2} \]

Here, \( Z \) is the atomic number which affects the energy level significantly in hydrogen-like atoms. The difference between these energy levels allows us to calculate the wavelength of light emitted or absorbed during an electron transition.
Understanding principal quantum numbers is crucial for interpreting the spectra of elements and extrapolating their physical characteristics.