Problem 4 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the Balmer series in the spectrum of \(\mathrm{H}\) atom, \(\bar{v}=R_{H}\left\\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\\}\), the correct statements among (I) to (IV) are: (I) As wavelength decreases, the lines in the series converge (II) The integer \(n_{1}\) is equal to 2 (III)The lines of longest wavelength corresponds to \(n_{2}=3\) (IV) The ionization energy of hydrogen can be calculated from wave number of these lines (a) (I), (III), (IV) (b) (I), (II), (III) (c) (I), (II), (IV) (d) (II), (III), (IV)

Step-by-Step Solution

Answer

The correct option is (b): (I), (II), (III).

Step 1: Understanding the Balmer Series Formula

The Balmer series in the hydrogen spectrum is given by \( \bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\} \). Here, \( \bar{v} \) is the wave number, \( R_H \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are integers such that \( n_2 \) > \( n_1 \). The Balmer series originates from transitions where electrons fall to the \( n_1 = 2 \) energy level.

Step 2: Evaluating Statement I

Statement (I) claims that as wavelength decreases, the lines in the series converge. In the Balmer series, as \( n_2 \) increases, the difference \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \) decreases, which leads to a decrease in wavelength because wave number \( \bar{v} \) increases. Therefore, as wavelength decreases, the lines indeed converge. Statement (I) is correct.

Step 3: Evaluating Statement II

Statement (II) states that the integer \( n_1 \) is equal to 2. This is indeed true for the Balmer series, as transitions end at the \( n_1 = 2 \) level. Hence, statement (II) is correct.

Step 4: Evaluating Statement III

Statement (III) indicates that the lines of longest wavelength correspond to \( n_2 = 3 \). The longest wavelength in a series corresponds to the smallest transition energy, which for the Balmer series is when \( n_2 = 3 \). Therefore, statement (III) is correct.

Step 5: Evaluating Statement IV

Statement (IV) claims that the ionization energy of hydrogen can be calculated from the wave number of these lines. However, ionization energy is typically derived from transitions from\( n = \infty \) to \( n_1 = 1 \) in the Lyman series, not from the Balmer series. Hence, statement (IV) is incorrect.

Step 6: Conclusion

The correct statements from (I) to (IV) are: (I), (II), and (III). Therefore, the correct option is (b): (I), (II), (III).

Key Concepts

Hydrogen spectrum

The hydrogen spectrum is a set of emission lines formed when light is emitted by hydrogen atoms transitioning between energy levels. When an electron in a hydrogen atom absorbs energy, it can move to a higher energy level. When it falls back, it emits light at specific wavelengths, creating a series of spectral lines. These lines are distinct for hydrogen and are primarily divided into several series, with the Balmer series being one of the most well-known.

Atomic spectra like the hydrogen spectrum provide evidence for the quantization of energy levels in atoms. This means electrons can only occupy specific energy levels, rather than a continuous range. This quantized nature is fundamental to the behavior and properties of atomic systems, and the hydrogen spectrum serves as a classic example of how quantum mechanics is applied to understand atomic emissions.

Wave number

Wave number is a measure of the number of waves per unit distance and is denoted as \( \bar{v} \). In the context of the hydrogen spectrum, it's a crucial concept that helps describe the position of spectral lines. The wave number is inversely related to wavelength, meaning shorter wavelengths correspond to larger wave numbers.

In the Balmer series formula, \( \bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\} \), wave number is used to calculate specific spectral transitions. As \( n_2 \) becomes larger, the difference \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \) gets smaller, increasing the wave number and thus decreasing the wavelength. This leads to the convergence of spectral lines as they approach the series limit.

Energy transitions

Energy transitions in hydrogen atoms occur when electrons move between quantized energy levels. When an electron drops from a higher energy level to a lower energy level, it emits a photon. The energy of this photon corresponds to the difference in energy between the two levels.

In the Balmer series, this transition specifically involves electrons moving down to the \( n_1 = 2 \) energy level. The energies associated with these transitions determine the wavelengths of light emitted. The relationship between energy levels and emitted light is key to understanding phenomena such as the formation of emission lines. The smallest energy transition in the Balmer series occurs when \( n_2 = 3 \), resulting in the longest wavelength of the series.

Rydberg constant

The Rydberg constant, denoted \( R_H \), is a fundamental constant in atomic physics used to predict the wavelengths of spectral lines of hydrogen. It appears in the formula for the Balmer series and is crucial for calculating wave numbers. It has a known value of approximately \( 1.097 \times 10^7 \) m\(^{-1}\).

Using the Rydberg constant, scientists and students can accurately determine the wave numbers of hydrogen spectral lines, thus telling us more about the specific energy transitions in the atom. This constant provides a universal tool for exploring atomic structure and processes, and it is an indispensable component in analyzing the hydrogen spectrum.