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The transition from the state \(n=4\) to \(n=3\) in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from: [2009] (A) \(3 \rightarrow 2\) (B) \(4 \rightarrow 2\) (C) \(5 \rightarrow 4\) (D) \(2 \rightarrow 1\)

Short Answer

Expert verified
The transition from \(5 \rightarrow 4\) (Option C) results in the emission of infrared radiation with a wavelength of 1950 nm, which falls within the infrared range of 700 nm to 1 mm.

Step by step solution

01

Rydberg Formula for Hydrogen-Like Atoms

The Rydberg formula for hydrogen-like atoms can be given by: \[ \frac{1}{\lambda} = R_\mathrm{H}Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \] where \(\lambda\) is the wavelength of the emitted radiation, \(R_\mathrm{H}\) is the Rydberg constant for hydrogen (\(1.097 \times 10^7\: \mathrm{m^{-1}}\)), \(Z\) is the atomic number (in this case, 1 for hydrogen), and \(n_1\) and \(n_2\) are the initial and final energy levels, respectively.
02

Calculate Wavelength for Each Transition

Using the given transitions in the options, we will calculate the wavelengths of the emitted radiation: (A) \(3 \rightarrow 2\): \[ \frac{1}{\lambda} = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \quad \Rightarrow \quad \lambda = 6.56\times10^{-7}\mathrm{m} \quad\text{(656 nm)} \] (B) \(4 \rightarrow 2\): \[ \frac{1}{\lambda} = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \quad \Rightarrow \quad \lambda = 4.86\times10^{-7}\mathrm{m} \quad\text{(486 nm)} \] (C) \(5 \rightarrow 4\): \[ \frac{1}{\lambda} = R_\mathrm{H}\left(\frac{1}{4^2} - \frac{1}{5^2}\right) \quad \Rightarrow \quad \lambda = 1.95\times10^{-6}\mathrm{m} \quad\text{(1950 nm)} \] (D) \(2 \rightarrow 1\): \[ \frac{1}{\lambda} = R_\mathrm{H}\left(\frac{1}{1^2} - \frac{1}{2^2}\right) \quad \Rightarrow \quad \lambda = 1.22\times10^{-7}\mathrm{m} \quad\text{(122 nm)} \]
03

Identify the Infrared Transition

Now that we have calculated the wavelengths for each transition, we can compare them to the wavelength range for infrared radiation (700 nm to 1 mm). From the calculated values, only the transition from option (C) \(5 \rightarrow 4\) has a wavelength within infrared range (1950 nm). It is therefore the correct answer: Transition from \(5 \rightarrow 4\) (Option C) results in the emission of infrared radiation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen-Like Atoms
Hydrogen-like atoms are atoms that resemble hydrogen in terms of their electron structure. They have only one electron orbiting the nucleus, similar to a hydrogen atom.
This single electron configuration leads to unique spectral lines which can be analyzed and predicted using specific formulas.
- **Atomic Number**: For hydrogen, the atomic number is 1, but for hydrogen-like ions such as He extsuperscript{+} or Li extsuperscript{2+}, the atomic number increases accordingly, affecting energy transitions. - **Spectral Lines**: Because these atoms have only one electron, the energy levels they can assume are quite distinct.
These levels are calculated using quantum mechanics, and the Rydberg formula helps in determining the position of spectral lines when transitions between levels occur.
The simplicity of hydrogen-like atoms makes them an excellent starting point for understanding more complex atoms.
Energy Transition
Energy transitions occur when an electron moves between different energy levels within an atom. This movement involves either absorbing or emitting energy.
This energy can be quantified, and the differences in energy levels lead to the emission or absorption of electromagnetic radiation, such as visible light or infrared radiation.
- **Quantized Levels**: Each level within an atom has a definite energy, and the transitions between these levels correspond to specific changes in energy. - **Electron Movement**: When an electron falls to a lower energy state, energy is emitted, often as light. Conversely, when it jumps to a higher state, the electron absorbs energy.
Understanding these transitions is crucial as they dictate the wavelength of radiation emitted or absorbed, which can be calculated using the Rydberg formula.
Wavelength Calculation
The wavelength of radiation during electron transitions in an atom is an essential aspect of understanding atomic spectra. The Rydberg formula allows us to calculate it by considering the specific energy levels involved.
The formula is:\[\frac{1}{\lambda} = R_\mathrm{H} Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right),\]where - \(\lambda\) is the wavelength,- \(R_\mathrm{H}\) is the Rydberg constant,- \(Z\) is the atomic number, and - \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels.
Using this formula, we can determine which transitions produce different types of radiation, such as visible light or infrared. Solving these calculations helps identify the specific position in the electromagnetic spectrum corresponding to the observed emission or absorption lines.
Infrared Radiation
Infrared radiation is a type of electromagnetic radiation found at wavelengths longer than visible light but shorter than microwave radiation. It is often associated with thermal energy.
- **Wavelength Range**: Infrared has a range from 700 nanometers (nm) to 1 millimeter (mm). - **Applications**: Commonly used in heaters, remote controls, and night-vision equipment due to its heat-emitting properties.
In atomic transitions, if the calculated wavelength falls into the infrared range, the transition is said to emit infrared radiation. For hydrogen-like atoms, specific transitions such as from higher quantum numbers to very close ones, like from 5 to 4, can result in infrared emissions. These phenomena are explored by observing how temperature and energy levels affect the infrared radiation characteristics.

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Most popular questions from this chapter

The bodies \(A\) and \(B\) have thermal emissivity's of \(0.01\) and \(0.81\) respectively. The outer surface areas of the two bodies are equal. The two bodies emit total radiant power at the same rate. The wavelength \(\lambda_{B}\) corresponding to maximum spectral radiance in the radiation from \(B\) is shifted from the wavelength corresponding to maximum spectral radiance in the radiation from \(A\), by \(1.00 \mu \mathrm{m}\). If the temperature of \(A\) is \(5802 \mathrm{~K}\). (A) the temperature of \(B\) is \(1934 \mathrm{~K}\). (B) \(\lambda_{B}=1.5 \mu \mathrm{m}\). (C) the temperature of \(B\) is \(1160 \mathrm{~K}\). (D) the temperature of \(B\) is \(2901 \mathrm{~K}\).

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