Problem 17 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Which of the following is the energy of a possible excited state of hydrogen ? (a) \(-3.4 \mathrm{eV}\) (b) \(+6.8 \mathrm{eV}\) (c) \(+13.6 \mathrm{eV}\) (d) \(-6.8 \mathrm{eV}\)

Step-by-Step Solution

Answer

The energy of a possible excited state of hydrogen is (a) \(-3.4 \mathrm{eV}\).

Step 1: Understand Energy Levels in Hydrogen Atom

The energy levels of a hydrogen atom are quantized and given by the formula: \( E_n = \frac{-13.6}{n^2} \) eV, where \(n\) is the principal quantum number (\(n=1,2,3,...\)). The ground state has \(n=1\), meaning the energy is \(-13.6\) eV. As \(n\) increases, the energy becomes less negative, reaching 0 eV as \(n\) approaches infinity.

Step 2: Determine Energy for Excited States

Excited states correspond to \(n > 1\). For example, for \(n = 2\), the energy is \( E_2 = \frac{-13.6}{2^2} = -3.4 \) eV. For \(n = 3\), \( E_3 = \frac{-13.6}{3^2} = -1.51 \) eV, and so on. Thus, excited state energies are always negative and greater than \(-13.6\) eV but less than 0 eV.

Step 3: Examine Given Energy Options

Review each option: - Option (a) \(-3.4 \mathrm{eV}\): This matches the energy of \(n = 2\), an excited state.- Option (b) \(+6.8 \mathrm{eV}\): Positive energy is not possible for bound states in hydrogen. - Option (c) \(+13.6 \mathrm{eV}\): Represents ionization energy, not an excited state. - Option (d) \(-6.8 \mathrm{eV}\): Does not correspond to any specific \(n\), and not an integer squared divisor of 13.6.

Step 4: Conclusion

From the above analysis, the energy of a possible excited state in hydrogen that matches an excited state energy level is option (a) \(-3.4 \mathrm{eV}\), corresponding to \(n=2\).

Key Concepts

Quantized Energy Levels

In the world of quantum mechanics, energy levels are not continuous but come in discrete steps. These are known as quantized energy levels. In the case of the hydrogen atom, the energy is described by \[ E_n = \frac{-13.6}{n^2} \text{ eV} \]where \(n\) is the principal quantum number. This formula shows how the energy levels depend on \(n\), which can only be a whole number (1, 2, 3,...).

• The ground state corresponds to the lowest energy level, where \(n = 1\).
• Higher energy levels are excited states, where \(n > 1\).
• The energies become less negative as \(n\) increases.

This quantization means that electrons in a hydrogen atom can't have just any energy; they must "jump" from one allowed energy state to another. This framework is crucial for understanding how electrons transition between different states and how spectra lines are formed.

Principal Quantum Number

The principal quantum number \(n\) is a fundamental component in quantum mechanics, especially in describing the electron configurations within an atom.

• It determines the energy level and size of an electron's orbit.
• As \(n\) increases, the electron's energy and orbital size also increase.
• A higher \(n\) value indicates that the electron is in a more excited, energetic state further from the nucleus.

In the case of hydrogen, \(n = 1\) is the ground state. Excited states occur at \(n = 2, 3, 4,\) and so on. The formula \( E_n = \frac{-13.6}{n^2} \, \text{eV} \) shows that energy is inversely proportional to the square of \(n\), explaining why higher values of \(n\) mean energy levels that are closer to zero.

Excited States

When an electron absorbs energy, it can be excited to a higher energy level, known as an excited state.

• Excited states occur for principal quantum numbers \(n > 1\).
• Each excited state is higher in energy compared to the ground state (\(n = 1\)).
• Typically, electrons don't stay long in these excited states and will eventually return to the ground state.
• As electrons return to lower energy levels, they release energy, often in the form of light.

For the hydrogen atom, only specific energies are possible in excited states. For example, the second energy level at \(n = 2\) has an energy of \(-3.4 \text{ eV}\).This precise quantization explains the specific lines seen in the hydrogen emission spectrum. In our example problem, the possible excited state for hydrogen is \(-3.4 \text{ eV}\), corresponding to \(n = 2\).