Problem 153 - A Complete Resource Book in Chemistry for JEE Main

Question

Energy of H-atom in the ground state is \(-3.6 \mathrm{eV}\), hence energy in the second excited state is (a) \(-6.8 \mathrm{eV}\) (b) \(-3.4 \mathrm{eV}\) (c) \(-1.51 \mathrm{eV}\) (d) \(-4.53 \mathrm{eV}\)

Step-by-Step Solution

Answer

The energy in the second excited state is \(-1.51 \, \text{eV}\), option (c).

Step 1: Understand the Energy Levels in the Hydrogen Atom

The energy levels of a hydrogen atom can be calculated using the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]where \(E_n\) is the energy of the \(n\)-th level and \(n\) is the principal quantum number. In this formula, \(-13.6 \, \text{eV}\) is the energy in the ground state \(n=1\).

Step 2: Identify the Second Excited State

The second excited state corresponds to the principal quantum number \(n = 3\) in the hydrogen atom. The ground state is \(n = 1\), the first excited state is \(n = 2\), and thus, the second excited state is \(n = 3\).

Step 3: Calculate the Energy of the Second Excited State

Substitute \(n = 3\) into the energy formula:\[E_3 = - \frac{13.6 \, \text{eV}}{3^2} = - \frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV}\]

Step 4: Compare the Calculation with the Options Given

The calculated energy for the second excited state is \(-1.51 \, \text{eV}\). Among the options provided, this matches option (c).

Key Concepts

Energy Levels Formula

In the world of hydrogen atoms, energy levels are quantized. This means electrons can only exist in distinct (or discrete) levels. An easy way to visualize this is to imagine a ladder. You can stand on the rungs, but not between them. In physics, we use a specific formula to calculate these energy levels:\[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]Here's what the formula means:- **\( E_n \)**: This represents the energy at a specific level, where n is greater than zero.- **\(-13.6 \text{ eV}\)**: This is the energy of the electron in the ground state (when it's at the lowest energy level, \(n=1\)).- **\(n\)**: The principal quantum number, which we'll discuss in the next section.The formula tells us that as \(n\) increases, the absolute value of the energy decreases (it becomes less negative), pointing to higher energy levels with less tightly bound electrons.

Principal Quantum Number

Every electron in a hydrogen atom is defined by a set of quantum numbers. The principal quantum number, represented by \(n\), is one of them. This number:- Determines the energy level of an electron.- Corresponds to larger orbitals when higher.For example:- **\(n = 1\)**: This is the ground state, the lowest energy level.- **\(n = 2\)**: Represents the first excited state.- **\(n = 3\)**: This is the second excited state and so on.Higher values of \(n\) imply electrons are further from the nucleus, increasing the potential energy. As a result, the energy associated with each level becomes less negative as \(n\) increases.

Second Excited State Calculation

Calculating the energy of a particular state in a hydrogen atom can be straightforward if we follow the steps. The formula, \( E_n = - \frac{13.6 \, \text{eV}}{n^2} \), will guide us through the process.Let's find the energy of the second excited state, which corresponds to the principal quantum number \(n = 3\):- **Step 1**: Substitute \(n = 3\) into the formula. \[ E_3 = - \frac{13.6 \, \text{eV}}{3^2} \]- **Step 2**: Simplify the expression: \[ E_3 = - \frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]Thus, the energy level for an electron in the second excited state of a hydrogen atom is \(-1.51 \, \text{eV}\). This matches option (c) from the exercise, confirming the calculations. Understanding these calculations is essential for grasping how electrons interact within an atom.