What is the frequency of revolution of electron present in 2 nd Bohr's orbit of \(H\) -atom? (a) \(1.016 \times 10^{16} \mathrm{~s}^{-1}\) (b) \(4.065 \times 10^{16} \mathrm{~s}^{-1}\) (c) \(1.626 \times 10^{15} \cdot \mathrm{s}^{-1}\) (d) \(8.2 \times 10^{14} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The frequency of revolution of an electron in the 2nd Bohr's orbit of a hydrogen atom is (b) \(4.065 \times 10^{16} \mathrm{~s}^{-1}\).

Step by step solution

01

Review Bohr's Model of the Hydrogen Atom

Recall that according to Bohr's model of the hydrogen atom, an electron moves in a circular orbit around the nucleus under the influence of the Coulomb force of attraction. The frequency of revolution of an electron in the nth orbit can be calculated using the expression for the radius of the nth orbit and the velocity of the electron in that orbit.
02

Express the Velocity of an Electron in the nth Orbit

The velocity of an electron in the nth Bohr orbit is given by the formula: \(v_n = \frac{e^2}{2h} \cdot \frac{1}{n}\), where \(e\) is the charge of an electron, \(h\) is Planck's constant, and \(n\) is the principal quantum number (the number of the orbit). For the 2nd orbit, \(n=2\).
03

Express the Frequency of Revolution for the 2nd Orbit

The frequency of revolution (\(f_n\)) is the number of revolutions per second, which can be expressed as the velocity of the electron divided by the circumference of the orbit: \(f_n = \frac{v_n}{2\pi r_n}\). The radius of the nth orbit is given by \(r_n = \frac{n^2h^2}{4\pi^2me^2}\), substituting for \(v_n\) and \(r_n\) in terms of \(n\), we can find an expression for frequency.
04

Calculate the Frequency for the 2nd Orbit

Using the equations from the previous steps, the frequency formula simplifies to \(f_n = \frac{me^4}{4h^3}\cdot \frac{1}{n^3}\). Plug in the known constants: mass of an electron \(m = 9.10938356 \times 10^{-31}\) kg, charge of an electron \(e = 1.602176634 \times 10^{-19}\) C, and Planck’s constant \(h = 6.62607015 \times 10^{-34}\) J·s. After substitution and calculation, find the frequency for the case of \(n=2\).
05

Select the Correct Answer

After performing the calculation, match the result with the given options to find the correct frequency of revolution for an electron in the 2nd Bohr's orbit of a hydrogen atom.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency of Electron Revolution
Understanding the frequency of electron revolution is essential when studying the atomic structure according to Bohr's model. In a hydrogen atom, electrons revolve around the nucleus in distinct orbits without radiating energy. These orbits are quantized, which means electrons can only occupy certain allowed paths with fixed energies.

The frequency of revolution, which is the rate at which an electron circles the nucleus in one of these paths, is inversely proportional to the cube of the principal quantum number (). This reflects how the electron’s speed decreases and the orbit radius increases as the electron moves to higher energy levels. Consequently, we see that an electron in the second orbit (n=2) revolves at a different frequency compared to the first orbit.

If the steps to calculate the frequency seems complex, breaking it down helps. The given formula in step 4 elegantly shows how the constants related to electron charge (), mass (), and Planck’s constant (), all play a part in determining this frequency. Using this formula, when an electron is in the second orbit, it completes fewer revolutions per second compared to the first orbit because the value of the principal quantum number () is squared in the equation's denominator.
Quantum Numbers
Quantum numbers are like the fingerprints of electrons; they describe the unique quantum state of an electron within an atom. There are four types of quantum numbers: principal (), angular momentum (), magnetic (), and spin (). When it comes to Bohr's model, the principal quantum number () is particularly significant as it defines the size of the electron's orbit and its energy level.

The principal quantum number, which can take any positive integer value, labels the orbits of an electron around a nucleus. For instance, in the 2nd orbit, the value for () is 2. This number not only impacts the radius of the electron's orbit and its energy level, but, as previously noted, it also influences the electron's frequency of revolution. The larger the quantum number, the farther the electron is from the nucleus and the more energy it possesses.

To put it simply, these quantum numbers are essential 'addresses' that give us detailed information about the probable location and energy of an electron in an atom, laying the groundwork for understanding the complex and fascinating behavior of these subatomic particles.
Coulomb Force
The Coulomb force, or electrostatic force, is the glue that holds the electrons in their orbits in Bohr’s model. It is the fundamental force that causes the attraction between the positively charged nucleus and the negatively charged electrons.

In the hydrogen atom, with only one proton and one electron, Coulomb's law describes the force of attraction as being directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This force is what counters the electron's inertia, which would otherwise cause it to fly off into space, and keeps it in a stable orbit.

It's the precise balance between this electrostatic force and the centripetal force required to keep an electron in a circular orbit that allows us to calculate the frequency of electron revolution as we saw in the step-by-step solution. Hence, without the Coulomb force, the essential structure of atoms as described by Bohr's model would not exist, and neither would the predictable frequencies of electron revolution which are pivotal to our understanding of atomic spectra.

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