Question

The transition the state \(\mathrm{n}=4\) to \(\mathrm{n}=1\) in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition form (A) \(3 \rightarrow 2\) (B) \(5 \rightarrow 4\) (C) \(4 \rightarrow 2\) (D) \(2 \rightarrow 1\)

Short answer

The correct transition that results in infrared radiation is (B) \(5 \rightarrow 4\), as it has a wavelength of approximately \(1875 \, \mathrm{nm}\), which falls within the infrared spectrum range.

Step-by-step solution

The Rydberg formula is given by \( \frac{1}{\lambda}=R_{\text{H}} \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \), where \(R_{\text{H}}\) is the Rydberg constant for hydrogen (\(1.097373\times10^7\, \text{m}^{-1}\)), \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels involved in the transition, and \( \lambda \) is the wavelength of the emitted radiation. The lower energy level has quantum number \(n_1\), and the higher energy level has quantum number \(n_2\). #Step 2: Calculate the wavelength for each transition#

We will now calculate the wavelength for each given transition using the Rydberg formula: (A) For the transition: \(3 \rightarrow 2\) \[ \frac{1}{\lambda_{A}} = R_{\text{H}} \left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \] (B) For the transition: \(5 \rightarrow 4\) \[ \frac{1}{\lambda_{B}} = R_{\text{H}} \left(\frac{1}{4^{2}}-\frac{1}{5^{2}}\right) \] (C) For the transition: \(4 \rightarrow 2\) \[ \frac{1}{\lambda_{C}} = R_{\text{H}} \left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right) \] (D) For the transition: \(2 \rightarrow 1\) \[ \frac{1}{\lambda_{D}} = R_{\text{H}} \left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \] #Step 3: Determine the wavelengths and check the infrared spectrum range#

Infrared radiation has wavelengths in the range of \(700 \, \mathrm{nm}\) to \(1 \, \mathrm{mm}\). After solving the equations in Step 2 for the wavelength, we can compare the values to the infrared range: (A) \( \lambda_{A} \approx 656 \, \mathrm{nm} \) (this is in the visible light spectrum) (B) \( \lambda_{B} \approx 1875 \, \mathrm{nm} \) (this is in the infrared spectrum) (C) \( \lambda_{C} \approx 486 \, \mathrm{nm} \) (this is in the visible light spectrum) (D) \( \lambda_{D} \approx 121 \, \mathrm{nm} \) (this is in the ultraviolet spectrum) #Step 4: Choose the correct option#

Among the given transitions, only the transition from \(5 \rightarrow 4\) results in a wavelength in the infrared spectrum. Therefore, the correct option is (B).

Key concepts

hydrogen atom

The hydrogen atom is the simplest and most well-known atom, consisting of just one proton and one electron. This simplicity makes it a fundamental subject of study in physics and chemistry.

• The proton, as a positively charged particle, resides in the nucleus.
• The electron, a negatively charged particle, orbits the nucleus in defined energy levels or shells.

These energy levels are quantized, meaning electrons can only occupy certain levels and not any energy level in between. When an electron transitions from a higher energy level to a lower one, it emits light of a specific wavelength. In the case of a transition from level 4 (n=4) to level 1 (n=1), the emitted light is part of the ultraviolet spectrum. This emission is because the energy difference between these levels is large. Understanding these transitions in a hydrogen atom is crucial for grasping the basics of quantum mechanics.

infrared radiation

Infrared radiation is a type of electromagnetic radiation that is invisible to the human eye but can be felt as heat. It occupies the spectrum just beyond visible red light, into longer wavelengths.

• Wavelength range: 700 nm to 1 mm
• Typically associated with heat and thermal emission

In the context of atomic transitions, such as those in the hydrogen atom, infrared radiation can occur when there is a small energy difference. For example, the transition from level 5 to level 4 in a hydrogen-like atom results in a wavelength around 1875 nm, which falls within the infrared range, as calculated using the Rydberg formula. Infrared radiation is not energetic enough to break chemical bonds but can excite molecules to higher vibrational states.

ultraviolet radiation

Ultraviolet (UV) radiation is a type of electromagnetic radiation that is more energetic than visible light, sitting in the spectrum just beyond violet light. It has the capability to cause chemical reactions and is known for its effects like sunburn.

• Wavelength range: 10 nm to 400 nm
• Can ionize atoms and molecules

In hydrogen atoms, transitions that involve a large change in principal quantum numbers, like from n=4 to n=1, release energy in the form of ultraviolet light. This is because the energy gap between these levels is significant enough to emit photons in the UV spectrum. Such transitions are used in UV spectroscopy to analyze chemical compositions and reactions.

principal quantum number

The principal quantum number, denoted as \(n\), is a fundamental concept in quantum mechanics. It determines the energy level of an electron within an atom.

• It is an integer value: 1, 2, 3, etc.
• The larger the \(n\), the higher the energy level and the further it is from the nucleus.

This number plays a crucial role in the calculation of the wavelengths of radiation emitted during electron transitions using the Rydberg formula. In scenarios like the one described, electrons transition between these quantized energy levels, and the difference in energy between initial (\(n_2\)) and final (\(n_1\)) levels determines the type of radiation emitted. Understanding the principal quantum number is key to predicting and explaining the various spectral lines observed in atomic spectra.