Problem 52 - Master Resource Book in JEE Main Physics

Question

The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is \(20,397 \mathrm{~cm}^{-1}\). The wave number of the energy for the same transition in \(\mathrm{He}^{+}\)is (a) \(5,099 \mathrm{~cm}^{-1}\) (b) \(20,497 \mathrm{~cm}^{-1}\) (c) \(14400 \mathrm{~A}\) (d) \(81,588 \mathrm{~cm}^{-1}\)

Step-by-Step Solution

Answer

The wave number is (d) 81,588 cm^{-1}.

Step 1: Understand the Rydberg Formula

The Rydberg formula for hydrogen-like atoms is given by \( \bar{u} = Z^2 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) where \( Z \) is the atomic number, \( R_H \) is the Rydberg constant for hydrogen, \( n_1 \) and \( n_2 \) are the principal quantum numbers corresponding to the final and initial orbits, respectively. For hydrogen, \( Z = 1 \). For \( \text{He}^+ \), \( Z = 2 \).

Step 2: Calculate the Energy for Hydrogen

Using the given wave number for hydrogen, \( \bar{u}_\text{H} = 20,397 \, \mathrm{cm}^{-1} \), we know it corresponds to the transition from the fourth orbit \( (n=4) \) to the second orbit \( (n=2) \) with \( Z = 1 \).

Step 3: Adjust for Helium Ion

Replace \( Z = 1 \) with \( Z = 2 \) in the Rydberg formula. The wave number for helium ion \( \text{He}^+ \) is \( \bar{u}_\text{He^+} = Z^2 \times \bar{u}_\text{H} \). Here, \( Z = 2 \), hence \( \bar{u}_\text{He^+} = 4 \times 20,397 = 81,588 \, \mathrm{cm}^{-1} \).

Step 4: Match the Result with Options

The calculated wave number for the same transition in \( \text{He}^+ \) is \( 81,588 \, \mathrm{cm}^{-1} \), which matches option (d).

Key Concepts

Wave Number

The concept of wave number is central to understanding atomic transitions. A wave number is defined as the number of wavelengths per unit distance and is commonly used in spectroscopy. It helps quantify the energy emitted or absorbed during electron transitions between orbits in an atom.
In mathematical terms, wave number \( \bar{u} \) is expressed in \( \text{cm}^{-1} \) and relates to energy through the equation \( E = h c \bar{u} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
Wave numbers provide a practical means of analyzing spectral lines, especially in understanding the quantized energy levels in hydrogen and hydrogen-like atoms. For these systems, the wave number directly corresponds to the difference in energy as electrons move from one orbit to another.

Quantum Numbers

Quantum numbers are essential in describing the properties and behaviors of electrons in atoms. Each electron in an atom can be characterized by a set of four quantum numbers. These numbers help identify the energy level, shape, and orientation of the electron's orbit, as well as its spin.
Here, we focus on the principal quantum number \( n \), which denotes the electron's energy level or orbit. It takes positive integer values (1, 2, 3, etc.). For the hydrogen atom transition discussed, the electrons move from the fourth orbit \( n_2 = 4 \) to the second orbit \( n_1 = 2 \).
Quantum numbers play a fundamental role in the Rydberg formula, which involves \( n_1 \) and \( n_2 \) to calculate the wavelengths or wave numbers during spectral transitions. This quantum mechanical framework explains how electrons transition between energy levels, leading to the emission or absorption of characteristic spectral lines.

Hydrogen-like Atoms

Hydrogen-like atoms are ions or atoms with only one electron, similar to the hydrogen atom. The simplest example is He+, a helium ion with one bound electron after losing its other electron. Such systems allow for easier analysis similar to hydrogen due to their one-electron structure.
The behavior of these atoms is important for understanding atomic structure and spectra. When using the Rydberg formula for hydrogen-like atoms, it's crucial to note the presence of the atomic number \( Z \). This accounts for the nuclear charge, affecting how tightly the electron is held.
For instance, in hydrogen, \( Z = 1 \). In the hydrogen-like ion He+, \( Z = 2 \). The Rydberg formula is modified by multiplying by \( Z^2 \) to accommodate this increased nuclear charge, leading to different wave lengths, or wave numbers, during electronic transitions. This allows us to accurately predict the energy involved in transitions such as the one in our original problem, resulting in a wave number of \( 81,588 \, \text{cm}^{-1} \) for the He+ ion, compared to hydrogen's \( 20,397 \, \text{cm}^{-1} \).