Question

If the shortest wavelength of \(\mathrm{H}\) atom in Lyman series is 'a', then longest wavelength in Balmer series of \(\mathrm{He}^{+}\) is: (a) \(\mathrm{a} / 4\) (b) \(5 \mathrm{a} / 9\) (c) \(4 \mathrm{a} / 9\) (d) \(9 \mathrm{a} / 5\)

Short answer

(d) \(9\mathrm{a}/5\)

Step-by-step solution

Understanding Lyman and Balmer Series

The Lyman series corresponds to electron transitions in a hydrogen atom from higher energy levels to the n=1 level, while the Balmer series relates to transitions to the n=2 level. The shortest wavelength occurs in the Lyman series when the transition is from n=infinity to n=1, while the longest wavelength in the Balmer series corresponds to transitions from n=3 to n=2.

Calculating Shortest Wavelength of Lyman Series

The formula for the wavelength of emitted light in the Lyman series is given by \( \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{n^2} \right) \) where \(R_H\) is the Rydberg constant for hydrogen. For the shortest wavelength (transition from \( n = \infty \) to \( n = 1 \)), the formula becomes \( \frac{1}{\lambda_{short}} = R_H \). Since the problem states this shortest wavelength is 'a', we have \( \lambda_{short} = \frac{1}{R_H} = a \).

Calculating Longest Wavelength of Balmer Series for \( \mathrm{He}^{+} \)

For the Balmer series, the wavelength is given by \( \frac{1}{\lambda} = R_{He^+} \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \) where \( R_{He^+} = Z^2 R_H \), with \( Z = 2 \) for \( He^+ \). For the longest wavelength (n=3 to n=2), it becomes \( \frac{1}{\lambda_{long}} = 4 R_H \left( \frac{1}{4} - \frac{1}{9} \right) \), which simplifies to \( \frac{1}{\lambda_{long}} = \frac{5}{36} R_H \). Solving for \( \lambda_{long} \), we get \( \lambda_{long} = \frac{36}{5 R_H} \).

Relating \( a \) to the Longest Wavelength

The longest wavelength in the Balmer series for \( He^+ \) is \( \lambda_{long} = \frac{36}{5 R_H} \), whereas the shortest wavelength in the Lyman series is \( a = \frac{1}{R_H} \). Thus, \( \lambda_{long} \) can be expressed in terms of \( a \) by \( \lambda_{long} = 36a/5 \). The options given in the problem can be rewritten, and it aligns with option (d): \( 9\mathrm{a}/5 \).

Key concepts

Hydrogen Atom Transitions

In the study of atomic physics, understanding how electrons transition between different energy levels in an atom is crucial. For hydrogen atoms, transitions are measured based on the energy levels of electrons. These transitions lead to the emission or absorption of light, resulting in spectral lines. Every line corresponds to a different transition.

• The Lyman series involves transitions ending at the first energy level ( =1), resulting in ultraviolet emissions.
• The Balmer series consists of transitions ending at the second energy level ( =2) and emits visible light.

These transitions are significant as they help us understand the intrinsic energetic properties of atoms. They also allow for the identification of elements present in stars and galaxies through spectral analysis.

Rydberg Formula

The Rydberg formula is an essential tool in atomic physics, predicting the wavelengths of spectral lines. It is applicable to hydrogen and hydrogen-like ions, such as the helium ion (He\(^+\)). The general form of the Rydberg equation is:
\[\frac{1}{\lambda} = RZ^2 \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)\]where:

• \( \lambda \) is the wavelength of the emitted or absorbed light.
• \( R \) is the Rydberg constant, specifically for hydrogen \( R_H \).
• \( Z \) is the atomic number of the element (1 for hydrogen, 2 for helium ion).
• \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels respectively.

The simplicity of the Rydberg formula makes it a powerful way to calculate and predict spectral lines. This makes it instrumental in deriving results such as the shortest and longest wavelengths in atom transition series.

Helium Ion He+ Transitions

Helium ions (He\(^+\)) can undergo transitions similar to the hydrogen atom, yielding different spectral series like the Balmer series. However, because of the different atomic number (\( Z = 2 \)), the spectral lines will be different from those of hydrogen.
In helium ions, transitions can be understood using the modified Rydberg formula, where the effective Rydberg constant becomes \( Z^2 R_H \), adding intensity to the transitions compared to hydrogen:
\[\frac{1}{\lambda} = 4R_H \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right)\]For example, in the Balmer series of \( He^+ \), the transition from =3 to =2 showcases the longest wavelength, calculated using this formula. These calculated values can provide insights into atomic structures and also play a role in astrophysics, aiding in the understanding of light patterns from space objects.