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For which atom or ion, the energy level of the second excited state is \(13.6 \mathrm{eV}\) ? (a) \(\mathrm{H}\) (b) \(\mathrm{He}^{+}\) (c) \(\mathrm{Li}^{2+}\) (d) Li

Short Answer

Expert verified
The ion for which the second excited state is \(13.6 \mathrm{eV}\) is the hydrogen ion \(\mathrm{H}^+\), as it is the only one with that energy for a second excited state.

Step by step solution

01

Understand the Energy Level Formula

The energy levels of an electron in a hydrogen-like atom or ion (that is, an atom or ion with only one electron, such as \(\mathrm{H}\), \(\mathrm{He}^{+}\), and \(\mathrm{Li}^{2+}\)) can be calculated using the formula \(E_n = \frac{-13.6 \mathrm{eV}}{n^2}\) where \(n\) is the principal quantum number corresponding to the energy level. The second excited state corresponds to \(n=3\) since the ground state is \(n=1\) and the first excited state is \(n=2\).
02

Apply the Energy Level Formula

Calculate the energy of the second excited state (\(n=3\)) using the formula for each given atom or ion. For hydrogen \(\mathrm{H}\), the energy is \(E_3 = \frac{-13.6 \mathrm{eV}}{3^2} = -1.51 \mathrm{eV}\). For helium ion \(\mathrm{He}^{+}\), the energy is scaled by the square of its atomic number \(Z=2\) as \(E_3 = \frac{-13.6 \mathrm{eV} \cdot Z^2}{3^2} = \frac{-13.6 \mathrm{eV} \cdot 4}{3^2} = -6.04 \mathrm{eV}\). Apply the method similarly to lithium ion \(\mathrm{Li}^{2+}\) and check for lithium atom \(\mathrm{Li}\), keeping in mind that the question implies a hydrogen-like atom or ion.
03

Identify the Correct Answer

Since we are looking for an energy level of \(13.6 \mathrm{eV}\), we need to find for which atom or ion the energy of the second excited state is positive due to the definition of excited states requiring energy input. By evaluating the previously calculated energies, we will identify the correct atom or ion.

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Key Concepts

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

Quantum Number
Quantum numbers are fundamental to understanding the structure and behavior of atoms, especially in atomic physics problems. They describe the unique state of an electron in an atom, which encompasses its energy, angular momentum, and magnetic properties. The most important quantum number in this context is the principal quantum number, designated as \( n \). It indicates the energy level or shell where an electron resides, with \( n=1 \) being the closest energy level to the nucleus and thus the lowest possible energy state, also known as the ground state.

When considering a hydrogen-like atom, the energy of an electron within any energy level can be calculated using the principal quantum number. Specifically, the formula \( E_n = \frac{-13.6 \text{eV}}{n^2} \) relates the energy of an electron to its principal quantum number, \( n \). This relationship is crucial for solving atomic physics problems, such as determining the energy levels of electrons in ions or atoms with a single electron in their orbital configuration.
Atomic Physics Problems
Solving atomic physics problems often involves applying the principles of quantum mechanics to predict the behavior of electrons in atoms. These problems might require calculations of energy levels, spectral lines, and probabilities of electron transitions. The example given in the original exercise pertains to calculating energy levels of hydrogen-like atoms, which is a cornerstone concept in atomic physics.

To tackle such problems, a strong conceptual understanding of the formulae and constants used in these calculations is essential. As seen in the exercise, knowing that the energy levels scale with the square of the atomic number \( Z \) for different ions allows us to determine the energy of an electron for any hydrogen-like ion. Subsequently, students are often asked to apply these concepts in competitive examinations in physical chemistry, where proficiency in swift and accurate problem-solving can greatly assist in securing high marks.
Physical Chemistry for Competitive Examinations
Success in competitive examinations for physical chemistry requires a deep understanding of various concepts, including the energy levels of hydrogen-like atoms. Questions on this topic not only test a student's knowledge of formulae but also their ability to apply concepts in diverse scenarios. For instance, these exams may include exercises where students need to find unknown variables, like determining the type of hydrogen-like ion based on its energy level.

Preparation for such examinations involves practising a large range of problems to gain familiarity with different types of questions and to develop a problem-solving strategy. To hit the mark, clear, concise explanations, as seen in step-by-step solutions, are invaluable. Such resources aid students in not only finding the answers but also understanding the underlying principles, ensuring a more comprehensive grasp of the material.

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Most popular questions from this chapter

When accelerated electrons are directed against an anticathode in an X-ray tube, the radiation obtained has a continuous spectrum with a wavelength minimum, \(\lambda_{\min }=\frac{1.24 \times 10^{-6}}{V} \mathrm{~m}\), where \(\mathrm{V}\) is the volt- age used for accelerating the electrons. \(\lambda_{\min }\) for \(V=5 \times 10^{4} \mathrm{~V}\) is (a) \(0.124 \mathrm{~nm}\) (b) \(0.248 \mathrm{~nm}\) (c) \(2.48 \mathrm{~nm}\) (d) \(1.24 \mathrm{~nm}\)

If \(n\) and \(l\) are, respectively, the principal and azimuthal quantum numbers, then the expression for calculating the total number of electrons in any energy level is (a) \(\sum_{l=1}^{l=n} 2(2 l+1)\) (b) \(\sum_{l=1}^{l=n-1} 2(2 l+1)\) (c) \(\sum_{l=0}^{l=n+1} 2(2 l+1)\) (d) \(\sum_{l=0}^{l=n-1} 2(2 l+1)\)

The ionization energy of H-atoms is \(13.6 \mathrm{eV}\). The ionization energy of deuterium atom should be (a) exactly \(13.6 \mathrm{eV}\) (b) slightly less than \(13.6 \mathrm{eV}\) (c) slightly greater than \(13.6 \mathrm{eV}\) (d) \(27.2 \mathrm{eV}\)

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) \(\frac{35}{36 R}\)

The wavelength of first line of Lyman series of \(\mathrm{H}\) -atom is \(1216 \AA\). What will be the wavelength of first line of Lyman series in 10 time ionized sodium atom \((Z=11)\) (a) \(1216 \AA\) (b) \(12.16 \AA\) (c) \(10 \AA\) (d) \(110 \AA\)

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