Question

When electrons are de-exciting to the ground state from \(n^{\text {th }}\) orbit of hydrogen atoms, 15 spectral lines are formed. The shortest wavelength among these will be (a) \(\frac{11}{900} R\) (b) \(\frac{900}{11 R}\) (c) \(\frac{35}{36} R\) (d) \(\frac{35}{36 R}\)

Short answer

The shortest wavelength among the spectral lines formed is \(\frac{36}{35R}\), which corresponds to option (d).

Step-by-step solution

Understanding the Spectral Lines Formula

The spectral lines formed when an electron moves between energy levels in a hydrogen atom are given by the formula for the number of spectral lines, which is \(N = \frac{n(n-1)}{2}\). Here, \(N\) is the number of spectral lines and \(n\) is the principal quantum number of the orbit the electron is de-exciting from.

Calculating the Principal Quantum Number

Given that there are 15 spectral lines, set up the equation \(15 = \frac{n(n-1)}{2}\) and solve for \(n\). Multiplying both sides by 2 gives us \(30 = n(n-1)\). Factoring or using the quadratic formula, we find that \(n = 6\) (since \(n\) must be positive).

Identifying the Transition for Shortest Wavelength

The shortest wavelength corresponds to the energy transition with the highest energy difference. This occurs when the electron moves from the highest energy level () to the ground state (\(n = 1\)).

Using the Rydberg Formula

Apply the Rydberg formula for the wavelength \(\lambda\) of the emitted photon during the transition: \(\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\), where \(R\) is the Rydberg constant, \(Z\) is the atomic number of hydrogen (which is 1), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.

Calculating the Shortest Wavelength

Insert the values for the transition from \(n = 6\) to \(n = 1\) into the Rydberg formula to find the shortest wavelength. Thus, \(\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{6^2}\right) = R\left(1 - \frac{1}{36}\right) = R\left(\frac{35}{36}\right)\). Therefore, the shortest wavelength is \(\lambda = \frac{36}{35}R^{-1} = \frac{36}{35R}\).

Key concepts

Rydberg Formula

The Rydberg Formula is crucial for understanding the spectral lines of a hydrogen atom. It is an equation that allows us to calculate the wavelengths of the photons emitted or absorbed when an electron transitions between energy levels in a hydrogen atom. The formula is expressed as:

\[\[\begin{align*}\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\end{align*}\]\]
Here, \( \lambda \) represents the wavelength of light, \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \) per meter), \( Z \) is the atomic number of the element, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively. In the context of the exercise, the Rydberg formula helped us determine the shortest wavelength, directly related to the most energetic transition, which occurs when an electron falls from the highest occupied energy level back to the ground state.

Principal Quantum Number

The Principal Quantum Number, symbolized by \( n \), is a crucial part of quantum mechanical modeling of an atom. It denotes the energy levels or 'orbits' that an electron can occupy around an atom's nucleus. Energy levels are quantized, which means electrons can only exist in specific orbits with definite energies, not in between them. In the exercise problem, we used the principal quantum number to determine the number of available electron transitions as an electron de-excites to the ground state, subsequently translating to the number of spectral lines observable. This number is integral because it provides insight into the electron configuration and dictates the electromagnetic radiation spectrum emitted by an atom.

Electron Transitions

Electron transitions refer to the movement of an electron between different energy levels within an atom, which occur when an electron gains or loses energy. This is the core activity behind the creation of spectral lines. In simple terms, when an electron jumps from a higher to a lower energy level, it releases energy in the form of a photon. The energy of this photon corresponds to the difference between the two energy levels. The shorter the wavelength of the emitted photon, the greater the energy released, which was a key point in solving our exercise. We identified that the shortest wavelength corresponds to the longest leap—it's all about how far the electron 'falls' that determines the wavelength of light produced.

Atomic Spectroscopy

Atomic spectroscopy is the study of the electromagnetic radiation absorbed and emitted by atoms. It is because each element emits light at characteristic wavelengths that this technique is a powerful tool for identifying elemental compositions and concentrations in samples. From a practical standpoint, by analyzing the spectral lines, as in our homework problem, one can determine various properties of an atom, such as its ionization states and the surrounding environment's influence on the atom. The patterns of these lines serve as unique signatures for elements, just like a fingerprint. As observed in the spectral lines emitted by excited hydrogen atoms, we can study the atomic spectra to gain insights into atomic structure and electron behaviors.