Frequency, Time Period And Angular Frequency

Introduction To Time-Period And Frequency Of Revolution Of An Electron

Chemistry in other words principally focuses on electron actions in an atom. With the help of Niels Bohr's atomic model, in which electrons were envisaged to revolve in circular orbits centred at the nucleus with fixed energy, it has provided a great advancement in understanding the atomic structure and chemical characteristics. Each specified by a principal quantum number is an energy state. Further, we get interested in the time period and frequency of an electron’s revolution in these orbits. Where, ‘time-period, ‘ varies as is the time taken by an electron to complete one full orbit around the nucleus, and ‘frequency, ‘, is the number of complete orbits per unit of time. These are important because they affect the energy state of electrons and the witnessed results during atomic emission and atomic absorption spectroscopy. Through these studies, chemists learn about transitions that occur to electrons, the energy that is absorbed or emitted, and an electronic configuration that determines the stability and reactivity of the atom.

In this article, we will cover the concept of Time Period and Frequency of the Revolution of an Electron in the n^th Bohr orbit. This concept falls under the broader category of Atomic structure, which is a crucial chapter in Class 11 chemistry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Let us study in detail the time Period and Frequency of Revolution of an Electron in the n^th Bohr orbit to gain insights into this topic and solve a few related problems.

Time Period and Frequency of Revolution of an Electron in the n^th Bohr orbit

As the spectroscopists, the time-period and frequency of an electron’s revolution in the nth Bohr orbit are basic necessary parameters in chemistry. Niels Bohr’s atomic model introduced the concept of orbits in which electrons revolve in circles around the nucleus and each orbit is defined by a number called the principal quantum number (n). The time-period (T_n) is the time taken by a particular electron to revolve in a particular orbit and can be determined as:

$\left(T_n=\frac{2 \pi r_n}{v_n}\right)$, where $\left(r_n\right)$ is the radius of the nth orbit and $\left(v_n\right)$ is the velocity of the electron in that orbit

On the other hand, frequency $\left[\nu_n\right]$ is defined as the number of vibrations per second or the number of complete oscillations per second, it is connected with the difference in energy levels between the concerned orbits. These concepts are very vital in spectroscopy, whereby they determine the emission and absorption of photons which correspond to energy transitions. They also determine the stability of the atomic nucleus as well as the chemical activity, giving details on the status of elements and molecules.

Although the precise equations for the time period and frequency of revolution are not required, it is still a good idea to look at the variations of these with the atomic number (Z) and the orbit number (n).

We know that Time period $(T)$ is the time required for one complete revolution and that Frequency $(\nu)$ is the inverse of the time period
$
\begin{aligned}
& \therefore T=\frac{\text { distance }}{\text { time }}=\frac{2 \pi r}{v} \\
& \because \mathrm{r} \propto \frac{\mathrm{n}^2}{\mathrm{Z}} \text { and } \mathrm{v} \propto \frac{\mathrm{Z}}{\mathrm{n}} \\
& \therefore T \propto\left(\frac{n^2}{Z} \times \frac{n}{Z}\right) \propto\left(\frac{n^3}{Z^2}\right) \\
& \therefore \nu=\left(\frac{1}{T}\right) \propto\left(\frac{Z^2}{n^3}\right)
\end{aligned}
$

You must remember all the above formulas and relations

For a better understanding of the topic and to learn more about Frequency, Time Period And Angular Frequency with video lesson we provide the link to the

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Solved Examples Based on Time Period and Frequency of Revolution of an Electron in the n^th Bohr Orbit

Example 1: Ratio of frequency of revolution of the electron in the $2^{\text {nd }}$ excited state of $H e^{+}$and $2^{\text {nd }}$ state state of hydrogen is.

1) (correct) $\frac{32}{27}$
2) $\frac{27}{32}$
3) $\frac{1}{54}$
4) $\frac{27}{2}$

Solution:

$
\begin{aligned}
& \frac{f_1}{f_2}=\frac{z_1^2}{n_1^3} \times \frac{n_2^3}{z_2^2} \\
& \Rightarrow n_1=3, n_2=3, z_1=2, z_2=1
\end{aligned}
$
$\therefore$ putting these value in the equation we get
$
\frac{2^2}{3^3} \times \frac{2^3}{1}=\frac{32}{27}
$

Hence, the answer is the option (1).

Example 2: The time taken for an electron to complete one revolution in the Bohr orbit of the hydrogen atom is

1) (correct) $\frac{4 \pi^2 m r^2}{n h}$
2) $\frac{n h}{4 \pi^2 m r}$
3) $\frac{n h}{4 \pi^2 m r^2}$
4) $\frac{h}{4 \pi m r}$

Solution:

We know that time period is the time taken for one complete revolution.

$\therefore T=\frac{2 \pi r}{V} \quad \ldots \ldots(1)$

Now, we know that the angular momentum of an electron in $n^{\text {th }}$ orbit is given by
$
m V r=\frac{n h}{2 \pi} \Rightarrow V=\frac{n h}{2 \pi m r} \ldots \ldots
$

From equations (1) and (2),
$
\therefore T=\frac{2 \pi r}{V}=\frac{2 \pi r}{\frac{n h}{2 \pi m r}}=\frac{4 \pi^2 m r^2}{n h}
$

Hence, the answer is the option (1).

Example 3:Determine the frequency of revolution of the electron in 3^rd Bohr's orbit in hydrogen atom:

1) (correct) 0.024 x 10¹⁶Hz

2) 3.4 x 10¹⁶Hz

3) 8.13 x 10¹⁶Hz

4) 0.054 x 10¹⁶Hz

Solution:

As we have learnt,

Frequency = 1/(Time period)

Period = Total distance covered/velocity

= 2πr/v

Thus, frequency = v/2πr

Now, velocity of electron in 3rd Bohr’s orbit = 2.16 x 10⁶(1/3)m/s

= 0.72 x 10⁶m/s

And, radius of 3rd Bohr’s orbit = 0.53 x 10^-10(3²/1)

= 4.77 x 10^-10m

Now, frequency = v/2πr

= 0.72 x 10⁶m/s / (2 x 3.14 x 4.77 x 10^-10m)

= 0.024 x 10¹⁶Hz

Hence, the answer is the option (1).

Example 4: The ratio of the orbital frequency of electron of hydrogen in the 3^rd and 2^nd orbital is:

1) 3.37

2) (correct)0.29

3) 0.44

4) 2.25

Solution

We know that,
Frequency $\propto \frac{\mathrm{Z}^2}{\mathrm{n}^3}$
For the same value of $Z$,
$
\begin{aligned}
& f \propto \frac{1}{n^3} \\
& \frac{f_3}{f_2}=\frac{\frac{1}{3^3}}{\frac{1}{2^3}}=\frac{8}{27}=0.29
\end{aligned}
$

Hence, the answer is the option (2).

Conclusion

Finally, It is comprehensible and quite reasonable that two of the most important concepts in atomic structure and atomic behaviour are the time period and frequency of revolution of the electrons in Bohr orbits. Niels Bohr’s model was essential in giving a foundation for a concept which stated that electrons orbit in a particular path with fixed energies. Regarding these orbits, two quantitative parameters known as the time period and the frequency affect atomic spectra, which show various lines for energy transitions. These parameters are basic in spectroscopy where the kinds of elements and composition of chemicals by emitted or absorbed radiation are determined. Also, they stress atomic stability, asserting that transitions between energy levels determine chemical activity and bonding characteristics. In addition to theoretical significance, knowledge about electron orbits improves technological advancements, like lasers and quantum computing, based on specific energy changes. Hence one can deduce an understanding of time-period and frequency concerning Bohr orbits that enhances understanding of atomic physics and stimulates improvement in fields of science as well as technology.