Problem 27 - University Physics with Modern Physics

Question

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n=1,2\) and 3 levels. (b) Calculate- the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is \(1.0 \times 10^{-8} \mathrm{s}\) . In the Bohr model, how many orbits does an electron in the \(n=2\) level complete before reurning to the ground level?

Step-by-Step Solution

Answer

(a) Speeds: \(2.18\times10^6\), \(1.09\times10^6\), \(7.27\times10^5\) m/s. (b) Periods: \(1.52\times10^{-16}\), \(6.08\times10^{-16}\), \(1.18\times10^{-15}\) s. (c) ~ \(1.64\times10^7\) orbits.

Step 1: Understand the Bohr Model Formulas

The Bohr model posits that the electron's speed in the nth orbit is given by \( v_n = \frac{k e^2}{\hbar n} \), where \( k \) is Coulomb's constant, \( e \) is the electron charge, \( \hbar \) is the reduced Planck's constant, and \( n \) is the principal quantum number. The orbital period \( T_n \) is \( T_n = \frac{2\pi r_n}{v_n} \), with \( r_n = \frac{n^2 \hbar^2}{k e^2 m_e} \), where \( m_e \) is the electron mass.

Step 2: Calculate the Electron Speed for Each Level

Substitute the known constants into the speed formula. Calculate \( v_1 \), \( v_2 \), and \( v_3 \):- For \( n=1 \), \( v_1 = \frac{2.18 \times 10^6}{1} \approx 2.18 \times 10^6 \) m/s.- For \( n=2 \), \( v_2 = \frac{2.18 \times 10^6}{2} \approx 1.09 \times 10^6 \) m/s.- For \( n=3 \), \( v_3 = \frac{2.18 \times 10^6}{3} \approx 7.27 \times 10^5 \) m/s.

Step 3: Calculate the Orbital Radius for Each Level

Using \( r_n = \frac{n^2 \hbar^2}{k e^2 m_e} \), determine the orbital radius:- For \( n=1 \), \( r_1 = 5.29 \times 10^{-11} \) m.- For \( n=2 \), \( r_2 = 4 \times 5.29 \times 10^{-11} \approx 2.12 \times 10^{-10} \) m.- For \( n=3 \), \( r_3 = 9 \times 5.29 \times 10^{-11} \approx 4.76 \times 10^{-10} \) m.

Step 4: Calculate Orbital Period for Each Level

Using \( T_n = \frac{2\pi r_n}{v_n} \), compute the period:- For \( n=1 \), \( T_1 \approx \frac{2\pi \times 5.29 \times 10^{-11}}{2.18 \times 10^6} \approx 1.52 \times 10^{-16} \) s.- For \( n=2 \), \( T_2 \approx \frac{2\pi \times 2.12 \times 10^{-10}}{1.09 \times 10^6} \approx 6.08 \times 10^{-16} \) s.- For \( n=3 \), \( T_3 \approx \frac{2\pi \times 4.76 \times 10^{-10}}{7.27 \times 10^5} \approx 1.18 \times 10^{-15} \) s.

Step 5: Calculate the Number of Orbits Before De-excitation

Using the result for the orbital period at \( n=2 \), divide the average lifetime by \( T_2 \):- Number of Orbits = \( \frac{1.0 \times 10^{-8}}{6.08 \times 10^{-16}} \approx 1.64 \times 10^7 \).

Key Concepts

Electron Speed

In the Bohr model, the electron speed within a hydrogen atom changes depending on the orbit it occupies. These orbits are defined by the quantum number \( n \). The formula for calculating the electron speed \( v_n \) in the \( n \)-th orbit is given by \( v_n = \frac{k e^2}{\hbar n} \):

• \( k \) represents Coulomb's constant.
• \( e \) is the electron's charge.
• \( \hbar \) denotes the reduced Planck's constant.
• \( n \) is the principal quantum number.

The speed decreases as \( n \) increases. This is because, in higher quantum levels, electrons are further from the nucleus and experience a weaker electrostatic pull. As a result: - For \( n=1 \), the speed is approximately \( 2.18 \times 10^6 \) m/s.- For \( n=2 \), it reduces to \( 1.09 \times 10^6 \) m/s.- For \( n=3 \), the speed is \( 7.27 \times 10^5 \) m/s. Understanding these speeds is crucial for appreciating how an electron moves across different energy states.

Orbital Period

The orbital period tells us how long it takes for an electron to complete one full orbit around the nucleus. Within the Bohr model, this period \( T_n \) is calculated using \( T_n = \frac{2\pi r_n}{v_n} \), where:

• \( r_n \) is the orbital radius.
• \( v_n \) is the electron speed.

As the quantum number \( n \) increases, both the orbital radius and period increase. Here's a quick look at the orbital periods at various levels: - For \( n=1 \), the orbital period is fleetingly small, just \( 1.52 \times 10^{-16} \) seconds. - At \( n=2 \), the period extends to \( 6.08 \times 10^{-16} \) seconds.- For \( n=3 \), it is about \( 1.18 \times 10^{-15} \) seconds. These periods, though minuscule, are a fantastic illustration of the rapidity with which electrons orbit within an atom.

Quantum Numbers

Quantum numbers are values that describe certain characteristics of electrons in atoms, including their energy level, shape, and orientation. In the Bohr model, the principal quantum number \( n \) is particularly significant as it determines the electron's orbit and energy level. Here's a breakdown of its main aspects:

• \( n \) signifies the energy level and orbit size. Higher \( n \) values indicate orbits with larger radii.
• \( n=1, 2, 3, ... \) signifies electrons with increasing energy, where \( n=1 \) is closest to the nucleus.
• Larger \( n \) values equate to lower speeds and longer orbital periods.

The quantum number \( n \) plays a vital role in predicting an electron's behaviour and helps physicists understand the structure of atoms.

Orbital Radius

The orbital radius \( r_n \) refers to the distance between the nucleus and the electron within a particular orbit. The Bohr model provides a fascinating way of calculating this distance with the formula \( r_n = \frac{n^2 \hbar^2}{k e^2 m_e} \). Here's a breakdown of what impacts the orbital radius:

• \( m_e \) is the electron mass.
• A higher \( n \) corresponds to a larger \( r_n \).

Let's see how the orbital radius varies: - For \( n=1 \), the orbital radius stands at \( 5.29 \times 10^{-11} \) meters. - For \( n=2 \), it increases to \( 2.12 \times 10^{-10} \) meters. - At \( n=3 \), the radius becomes \( 4.76 \times 10^{-10} \) meters. The progressive increase in \( r_n \) with \( n \) demonstrates how electrons occupy larger spaces as they absorb more energy.