Problem 56 - Master Resource Book in JEE Main Physics

Question

The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number \(n\) are \(E, P\) and \(r\) respectively. Then according to Bohr's theory of hydrogen atom, (a) \(P r\) is proportional to \(n\) (b) \(P / E\) is proportional to \(n\) (c) \(E r\) is constant for all orbits (d) \(E P r\) is proportional to \(1 / n\)

Step-by-Step Solution

Answer

Options (a), (c), and (d) are correct.

Step 1: Understand Bohr's Model Equations

Bohr's model describes that the total energy of an electron in a hydrogen orbit is given by \( E_n = -\frac{kZ^2e^4m}{2\hbar^2n^2} \), where \( k \) is Coulomb's constant, \( Z \) is the atomic number, \( e \) is the electron charge, \( m \) is the electron mass, \( \hbar \) is the reduced Planck's constant, and \( n \) is the quantum number. Also, the radius of the orbit is \( r_n = \frac{n^2\hbar^2}{kZe^2m} \). The linear momentum \( P \) of the electron is given by \( P = m v \), where \( v \) is the velocity.

Step 2: Connect Momentum and Radius with Quantities

For the linear momentum \( P \), according to Bohr's model, the velocity \( v \) is \( v = \frac{Z e^2}{n\hbar} \). So, \( P = m v = \frac{mZe^2}{n\hbar} \). Now, substituting \( r \) and \( P \) values into expressions: \( P r = \frac{mZe^2}{n\hbar} \cdot \frac{n^2\hbar^2}{kZe^2m} = \frac{n\hbar}{k} \), which shows \( P r \) is proportional to \( n \). Hence, option (a) is correct.

Step 3: Check Other Proportionalities

For \( \frac{P}{E} \), substitute \( P \) and \( E \): \( \frac{P}{E} = \frac{\frac{mZe^2}{n\hbar}}{-\frac{kZ^2e^4m}{2\hbar^2n^2}} = -\frac{2n\hbar}{kZ e^2} \). This shows \( \frac{P}{E} \) is not proportional to \( n \), rather \( n^{-1} \). Next, for \( Er \), substitute \( E \) and \( r \): \( E r = -\frac{kZ^2e^4m}{2\hbar^2n^2} \times \frac{n^2\hbar^2}{kZe^2m} = -\frac{Ze^2}{2} \), a constant. Thus option (c) is correct. Lastly, for \( EPr \): \( EPr = E \times P \times r = -\frac{kZ^2e^4m}{2\hbar^2n^2} \times \frac{mZe^2}{n\hbar} \times \frac{n^2\hbar^2}{kZe^2m} \), simplifying gives \( EPr \propto n^{-1} \). Thus option (d) is correct too.

Step 4: Conclusion about Proportionalities

We found that \( P r \) is proportional to \( n \); \( E r \) is constant for all orbits; and \( E P r \) is proportional to \( 1/n \). Hence, options (a), (c), and (d) are correct.

Key Concepts

Energy Levels

In Bohr's model of the hydrogen atom, energy levels are like distinct steps that an electron can occupy. Each energy level is defined by a principal quantum number, denoted as \( n \). This quantum number can take integer values such as 1, 2, 3, and so on. The energy of an electron becomes more negative as \( n \) decreases, meaning that the electron is more tightly bound to the nucleus in lower energy levels.

Mathematically, the energy \( E_n \) of an electron in a particular orbit or energy level is expressed as:

• \( E_n = -\frac{kZ^2e^4m}{2\hbar^2n^2} \)

This formula incorporates constants like Coulomb's constant \( k \), and the reduced Planck's constant \( \hbar \). The energy is inversely proportional to the square of \( n \), indicating that higher quantum numbers mean higher (less negative) energies.

Angular Momentum

Angular momentum in Bohr's theory provides insight into how electrons behave as they orbit the nucleus. It is quantized, meaning it can only take specific discrete values, a concept that was revolutionary at the time.

According to Bohr, the angular momentum \( L \) of an electron is given by:

• \( L = n \hbar \)

where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant. This equation signifies that angular momentum increases linearly with \( n \).

Quantization of angular momentum implies that electrons can only orbit at certain distances from the nucleus, corresponding to specific allowed energy states. This was a key element in explaining why electrons do not spiral into the nucleus, as was a challenge in the earlier atomic models.

Quantum Number Effects

Quantum numbers play a pivotal role in determining the properties and behavior of electrons within atoms. They are like the ID numbers for electrons' positions and energy levels within an atom.

For an electron in a hydrogen atom, the principal quantum number \( n \) determines:

• The energy level or shell where the electron resides.
• The size of the electron's orbit, as higher \( n \) means larger orbits.
• The electron's energy, with energy increasing as \( n \) increases.

Other quantum numbers (not focused on in Bohr's model) further define electron properties, such as angular momentum's orientation and spin. Understanding these effects is essential for explaining chemical bonding and spectral lines observed in experiments.

Electron Orbits

Electron orbits in Bohr's model are circular paths that electrons follow around the nucleus. These orbits are stationary, meaning that electrons in these orbits do not radiate energy unless they change orbits.

The radius of an orbit relates to the principal quantum number \( n \) by the equation:

• \( r_n = \frac{n^2\hbar^2}{kZe^2m} \)

This reveals that the radius depends on \( n^2 \), making higher energy levels significantly larger due to their \( n^2 \) dependency.

Per Bohr's model, when an electron jumps from one orbit to another, it emits or absorbs a photon with an energy equal to the difference between the two orbits. This quantization of orbits was crucial for explaining the discrete spectral lines observed in atomic emission and absorption spectra. This revolutionized understanding of atomic structure, leading the way towards modern quantum theory.