Question

The wave number of first line in the Balmer series of hydrogen is \(15200 \mathrm{~cm}^{\text {? }}\). The wave number of the first line in the Balmer series of \(\mathrm{Be}^{3+}\) is (1) \(2.43 \times 10^{5} \mathrm{~cm}^{-1}\) (2) \(3.43 \times 10^{5} \mathrm{~cm}^{-1}\) (3) \(4.43 \times 10^{5} \mathrm{~cm}^{-1}\) (4) \(5.43 \times 10^{5} \mathrm{~cm}^{-1}\)

Short answer

The wave number for \(\mathrm{Be}^{3+}\) is \(2.43 \times 10^5 \mathrm{~cm}^{-1}\).

Step-by-step solution

Understand the formula for wave number

Wave number is given by the formula: \[ \tilde{v} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]. Here, \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the states involved in the transition.

Identify values for hydrogen

For hydrogen in the Balmer series: \[ n_1 = 2 \text{ and } n_2 = 3 \]. The atomic number \(Z\) for hydrogen is 1. We are given the wave number for the first line as \(15200 \mathrm{~cm}^{-1}\).

Relate wave numbers for different elements

The wave number for another element can be calculated if we know its atomic number. The atomic number for \( \mathrm{Be}^{3+} \) is 4. Thus, we can write the relation: \[ \tilde{v}_{ \mathrm{Be}^{3+} } = \tilde{v}_{ \mathrm{H} } \left( \frac{Z_{ \mathrm{Be}^{3+} }^2}{Z_{ \mathrm{H} }^2} \right) \].

Calculate wave number for \( \mathrm{Be}^{3+} \)

Substitute the known values into the relation: \[ \tilde{v}_{ \mathrm{Be}^{3+} } = 15200 \times \left( \frac{4^2}{1^2} \right) \].

Simplify and solve

Simplify the expression: \[ \tilde{v}_{ \mathrm{Be}^{3+} } = 15200 \times 16 = 243200 \mathrm{~cm}^{-1} \].

Key concepts

Wave Number

The wave number, often represented by the symbol \(\tilde{v}\), is a fundamental concept in spectroscopy that refers to the number of wavelengths per unit distance. It is inversely related to the wavelength, \( \tilde{v} = \frac{1}{\text{wavelength}} \), making it very useful in describing transitions in atomic spectra. When a photon is emitted or absorbed, the wave number helps to determine the energy changes associated with that transition. Its standard unit is \( \text{cm}^{-1} \). Understanding wave numbers allows us to relate physical measurements to quantum mechanical calculations efficiently. For example, in the Balmer series of hydrogen, the given wave number is 15200 \( \text{cm}^{-1} \), which describes the first line where an electron transitions between energy levels.

Hydrogen Spectrum

The hydrogen spectrum arises from electrons transitioning between various energy levels within a hydrogen atom. Specifically, the Balmer series is a significant part of the hydrogen spectrum observed in the visible region. When an electron drops from a higher energy level (\( n_2 \)) to a lower energy level (\( n_1 = 2 \)), it emits a photon whose wave number can be calculated using the Rydberg formula. This helps identify specific lines in the spectrum. Each line corresponds to a particular electronic transition. For instance, the first line in the Balmer series involves an electron descending from \( n_2 = 3 \) to \( n_1 = 2 \). This results in a specific wavelength or wave number characteristic to hydrogen.

Atomic Transitions

Atomic transitions occur when an electron moves between energy levels within an atom. The transition between these levels results in either the absorption or emission of light at specific wavelengths, forming spectral lines. The energy difference between the initial and final states of the electron determines the wavelength or wave number of the emitted or absorbed photon. For the Balmer series, these transitions end at \( n_1 = 2 \). This is why the observed lines are only in the visible spectrum. The formula \( \tilde{v} = RZ^2 \bigg( \frac{1}{n_1^2} - \frac{1}{n_2^2} \bigg) \) helps in calculating the wave number of the spectral lines efficiently.

Rydberg Constant

The Rydberg constant, denoted by \( R \), is a fundamental constant in atomic physics that appears in the Rydberg formula. It quantifies the relationship between wavelength (or wave number) of emitted or absorbed photons during electron transitions. For hydrogen, the Rydberg constant is approximately \( 1.097 \times 10^7 \) \( \text{m}^{-1} \). In the context of our problem, using the Rydberg constant helps us relate the wave numbers for hydrogen to other elements like \( \text{Be}^{3+} \). By knowing the value of \( Z \) for the given transitions, we can predict the spectral lines for any hydrogen-like ion. For \( \text{Be}^{3+} \) with \( Z = 4 \), the significant factor is the ratio of \( Z^2 \) between the species, enhancing understanding of atomic spectra across different elements.