What Transition In The Hydrogen Atom Corresponds To The Same Emitted Light Frequency As The Transition From N=4 To N=2 In He+?

In the realm of atomic physics, understanding the frequencies of light emitted during electronic transitions is paramount. When an electron transitions between energy levels within an atom or ion, it emits or absorbs a photon with energy equal to the energy difference between the levels. This phenomenon is governed by fundamental principles of quantum mechanics, and the specific frequencies emitted are unique to each element and its ionic state. In this article, we will delve into the transition of an electron from the $n=4$ to $n=2$ energy level in the helium ion ($He^{+}$) and compare it to transitions in the hydrogen atom (H). Our focus will be on identifying the hydrogen atom transition that corresponds to the same emitted light frequency as the helium ion transition. This exploration will provide insights into the energy level structure of these atomic systems and the relationship between atomic number and transition energies.

Key Concepts and Principles

Before diving into the specifics, it's important to grasp the underlying principles that govern electronic transitions and light emission. The energy levels in an atom are quantized, meaning electrons can only occupy specific energy states. These energy levels are described by the principal quantum number, n, which can take integer values (1, 2, 3, and so on), with higher numbers indicating higher energy levels. When an electron transitions from a higher energy level ($n_i$) to a lower energy level ($n_f$), it releases energy in the form of a photon. The energy (E) of this photon is precisely the difference between the initial and final energy levels:

$E = E_i - E_f$

This energy is also related to the frequency (ν) and wavelength (λ) of the emitted light through Planck's equation:

$E = hν = hc/λ$

where h is Planck's constant and c is the speed of light. The Rydberg formula provides a mathematical framework for calculating the wavelengths and frequencies of light emitted during electronic transitions in hydrogen-like species (species with only one electron). The formula is given by:

$1/λ = RZ^2 (1/n_f^2 - 1/n_i^2)$

where R is the Rydberg constant, Z is the atomic number (number of protons), $n_i$ is the initial energy level, and $n_f$ is the final energy level. This formula is crucial for comparing transitions in different atoms and ions.

Calculating the Frequency for $He^{+}$ Transition ($n=4$ to $n=2$)

To determine the frequency of light emitted during the transition from $n=4$ to $n=2$ in $He^{+}$, we will employ the Rydberg formula. For $He^{+}$, the atomic number Z is 2. Plugging in the values, we get:

$1/λ = R(2^2) (1/2^2 - 1/4^2) = 4R (1/4 - 1/16) = 4R (4/16 - 1/16) = 4R (3/16) = (3/4)R$

This equation gives us the reciprocal of the wavelength. To find the frequency (ν), we use the relationship:

$ν = c/λ$

So,

$ν = c(3/4)R = (3/4)cR$

This is the frequency of light emitted when an electron transitions from the $n=4$ to $n=2$ level in $He^{+}$. Now, we need to identify which transition in the hydrogen atom will produce the same frequency.

Determining the Equivalent Transition in Hydrogen Atom

For the hydrogen atom, the atomic number Z is 1. We need to find a transition ($n_i$ to $n_f$) in hydrogen such that its emitted frequency matches the one we calculated for $He^{+}$. The Rydberg formula for hydrogen is:

$1/λ = R(1^2) (1/n_f^2 - 1/n_i^2) = R (1/n_f^2 - 1/n_i^2)$

And the frequency is:

$ν = c/λ = cR (1/n_f^2 - 1/n_i^2)$

We want this frequency to be equal to the frequency of the $He^{+}$ transition, which is $(3/4)cR$. Therefore, we set the two frequencies equal to each other:

$cR (1/n_f^2 - 1/n_i^2) = (3/4)cR$

Dividing both sides by cR, we get:

$1/n_f^2 - 1/n_i^2 = 3/4$

Now, we need to find integer values for $n_i$ and $n_f$ that satisfy this equation. Let's examine the provided options to see which one fits.

Analyzing the Options

• Option A: $n=2$ to $n=1$ For this transition, $n_i = 2$ and $n_f = 1$. Plugging these values into the equation:

  $1/1^2 - 1/2^2 = 1 - 1/4 = 3/4$

  This matches the required value of $3/4$. Therefore, the transition from $n=2$ to $n=1$ in the hydrogen atom has the same emitted frequency as the transition from $n=4$ to $n=2$ in the helium ion.

• Option B: $n=3$ to $n=2$ For this transition, $n_i = 3$ and $n_f = 2$. Plugging these values into the equation:

  $1/2^2 - 1/3^2 = 1/4 - 1/9 = (9 - 4)/36 = 5/36$

  This does not equal $3/4$, so this option is incorrect.

Conclusion

In summary, the frequency of light emitted during the transition from $n=4$ to $n=2$ in the helium ion ($He^{+}$) is equal to the frequency of light emitted during the transition from n=2 to n=1 in the hydrogen atom (H). This conclusion is reached by applying the Rydberg formula and comparing the energy differences between the electronic levels in both species. The Rydberg formula, a cornerstone of atomic physics, allows us to quantitatively analyze and predict the wavelengths and frequencies of light emitted during electronic transitions in hydrogen-like species. By understanding the interplay between the Rydberg constant, atomic number, and principal quantum numbers, we can gain valuable insights into the electronic structure of atoms and ions. The helium ion transition from n=4 to n=2 exemplifies how transitions in heavier, ionized species can correspond to specific transitions in simpler atoms like hydrogen. This correlation arises from the increased nuclear charge in the helium ion, which alters the energy level spacing compared to hydrogen. This analysis underscores the fundamental principles governing atomic spectra and provides a foundation for understanding more complex atomic systems. The frequency of light emitted is directly proportional to the energy difference between the initial and final states, highlighting the quantized nature of electron energy levels within atoms. This concept is crucial for various applications, including spectroscopy, where the unique spectral fingerprints of elements are used for identification and analysis. Furthermore, understanding electronic transitions is essential in fields such as astrophysics, where the light emitted by distant stars and galaxies provides information about their composition and physical conditions. The implications of these principles extend to technological applications as well, including the development of lasers, lighting technologies, and various analytical techniques. By meticulously comparing the transitions in helium ion and hydrogen atom, we reinforced the applicability and importance of the Rydberg formula. This formula is not only a theoretical construct but also a powerful tool for predicting and interpreting experimental observations. The accurate calculation of transition frequencies is essential for various spectroscopic techniques, allowing scientists to identify elements and study their electronic structures with high precision. This article serves as a concise yet comprehensive exploration of the relationship between electronic transitions and light emission, demonstrating the power of quantum mechanical principles in explaining the behavior of atoms and ions. The comparison between helium ion and hydrogen atom transitions provides a clear illustration of how the atomic number affects the energy levels and spectral properties of these species. By mastering these fundamental concepts, students and researchers can delve deeper into the fascinating world of atomic physics and its myriad applications. Further exploration might involve studying transitions in other elements, considering the effects of electron spin and relativistic corrections, and investigating the role of external fields on atomic spectra. The journey into atomic physics begins with understanding these basic principles, and the exploration of electronic transitions is a key step in this exciting scientific endeavor.