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\(^{-8}\) s. In the Bohr model, how many orbits does an electron in the \(n\) = 2 level complete before returning to the ground level?

Short answer

The speeds are 2.18, 1.09, and 0.73 million m/s for n=1, 2, 3. Periods are 1.52x10^-16, 1.22x10^-15, and 4.10x10^-15 s. ~8.2 million orbits before dropping to n=1.

Step-by-step solution

Understanding the Bohr Model

The Bohr model provides a way to calculate the speed of an electron in an orbit around the nucleus of a hydrogen atom. The formula for the speed is given by \( v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \), where \( e \) is the electron charge, \( \varepsilon_0 \) is the permittivity of free space, \( h \) is Planck's constant, and \( n \) is the principal quantum number.

Calculating Speed for Different Energy Levels

Substitute the known constants and calculate the speed for each energy level \( n = 1, 2, 3 \):- \( n = 1 \): \( v_1 = 2.18 \times 10^6 \) m/s- \( n = 2 \): \( v_2 = \frac{v_1}{2} = 1.09 \times 10^6 \) m/s- \( n = 3 \): \( v_3 = \frac{v_1}{3} = 0.73 \times 10^6 \) m/s

Formula for Orbital Period

The orbital period \( T \) can be calculated using the formula \( T = \frac{2\pi r_n}{v_n} \), where \( r_n = n^2 a_0 \) with \( a_0 \) as the Bohr radius \( 5.29 \times 10^{-11} \) m. This gives an expression for the period as \( T_n = \frac{2\pi n^3 \cdot a_0}{v_1} \).

Calculating Orbital Period for Each Level

Calculate \( T_n \) for levels \( n = 1, 2, 3 \):- \( n = 1 \): \( T_1 = \frac{2\pi \cdot 5.29 \times 10^{-11}}{2.18 \times 10^6} \approx 1.52 \times 10^{-16} \) s- \( n = 2 \): \( T_2 = 2^3 \cdot T_1 = 1.22 \times 10^{-15} \) s- \( n = 3 \): \( T_3 = 3^3 \cdot T_1 = 4.10 \times 10^{-15} \) s

Orbits in First Excited Level

Calculate the number of orbits in \( n=2 \) before returning to \( n=1 \). The number of orbits \( N \) is given by \( N = \frac{L}{T} \) where \( L = 1.0 \times 10^{-8} \) s and \( T = 1.22 \times 10^{-15} \) s from Step 4. Compute \( N = \frac{1.0 \times 10^{-8}}{1.22 \times 10^{-15}} \approx 8.2 \times 10^6 \) orbits.

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest atom and plays a critical role in understanding broader atomic structure in physics. It consists of a single proton as the nucleus around which a single electron orbits. In the Bohr model, this setup provides a way to analyze the electron's energy levels and behavior when it transitions between them.
The Bohr model was revolutionary because it introduced the idea of quantized energy levels, meaning the electron can only occupy certain orbits with specific energy. This concept was integral in explaining the emission spectra of hydrogen, which are the light waves emitted when its electron transitions from a higher energy level to a lower one.
In essence, the hydrogen atom allows physicists to test theories about atomic energy levels and electron dynamics in a controlled and predictable manner, providing a foundation for more complex atomic models.

Principal Quantum Number

The principal quantum number, denoted by \( n \), is a key component of quantum mechanics and the Bohr model, helping to describe the energy levels of an electron in an atom. In terms of the hydrogen atom, the principal quantum number determines the possible orbits an electron can inhabit.

• For \( n = 1 \), the electron occupies the closest orbit to the nucleus, known as the ground state.
• As \( n \) increases (e.g., \( n = 2, 3 \)), the electron resides in higher energy orbits, further from the nucleus.

Each principal quantum number corresponds to a different amount of energy, with higher values of \( n \) indicating higher energy levels due to increased distance from the nucleus. This quantum number is crucial for calculating other properties like the electron's speed and the orbital period, providing insight into how electrons behave within atoms.

Orbital Period

The orbital period refers to the time it takes for the electron to complete one full orbit around the nucleus. In the Bohr model for the hydrogen atom, the orbital period depends on the principal quantum number \( n \) and can be determined using the formula:
\[T_n = \frac{2\pi n^3 \cdot a_0}{v_1}\]
Where \( a_0 \) is the Bohr radius, and \( v_1 \) is the speed of the electron at the ground state.
For different levels:

• At \( n = 1 \), the orbital period is very short, about \( 1.52 \times 10^{-16} \) seconds.
• Higher levels have longer periods; for example, \( n = 2 \) has \( T_2 = 1.22 \times 10^{-15} \) seconds.
• An even higher level like \( n = 3 \) results in a period of \( T_3 = 4.10 \times 10^{-15} \) seconds.

The period increases with higher \( n \) because the electron travels a larger orbit at slower speeds. This concept helps quantify how energetic transitions occur in atoms.

Electron Speed

In the Bohr model of the hydrogen atom, the speed of an electron is dictated by its orbit or principal quantum number. The speed decreases as the electron's orbit becomes larger, or as \( n \) increases.
The speed \( v_n \) of an electron in level \( n \) is given by:
\[v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n}\]
Where \( e \) is the electron charge, \( \varepsilon_0 \) is the permittivity of free space, and \( h \) is Planck's constant.

• At \( n = 1 \), the electron's speed is around \( 2.18 \times 10^6 \) m/s.
• For \( n = 2 \), it slows to approximately half, or \( 1.09 \times 10^6 \) m/s.
• At \( n = 3 \), the speed further decreases to \( 0.73 \times 10^6 \) m/s.

This relationship shows how an electron's dynamics change with its energy level, affecting the atom's interaction with light and other particles. Understanding this electron speed is essential for deeper studies in quantum mechanics and atomic physics.