If an electron is moving in the nth orbit of the hydrogen atom, then its velocity v_n for the nth or

Q: Why does the electron slow down in higher orbits?

As the electron moves to higher orbits, it gets further from the nucleus. The electrostatic force of attraction weakens, meaning less centripetal force is required to keep it in a circular path. Consequently, it doesn't need to move as fast.

Q: Do I need to derive this formula during the exam?

No, deriving it takes too much time. It is highly recommended to memorize the proportionalities: $v_n \propto 1/n$, $r_n \propto n^2$, and $E_n \propto 1/n^2$.

Q: What is the velocity in the first orbit of hydrogen?

The velocity in the first orbit ($n=1$) is approximately $2.18 \times 10^6 \text{ m/s}$. This is often written as $c/137$, where $c$ is the speed of light.

Q: Does this proportionality hold for ions like He+ or Li2+?

Yes! For hydrogen-like single-electron ions, the velocity is $v_n \propto Z/n$. Since $Z$ is constant for a specific ion, the velocity is still inversely proportional to $n$.

Q: Can the velocity ever reach zero?

According to the formula, velocity approaches zero only as $n$ approaches infinity. Physically, this means the electron is completely removed from the atom, and the atom is ionized.

Question

If an electron is moving in the $n$th orbit of the hydrogen atom, then its velocity $v_n$ for the $n$th orbit is given as

Options:

(A) $v_n \propto n$

(B) $v_n \propto \frac{1}{n}$

(C) $v_n \propto n^2$

(D) $v_n \propto \frac{1}{n^2}$

Sigma Percentile · Accepted Answer

### The Cosmic Dance of the Electron
Imagine you are standing at the center of a hydrogen atom. Right beside you is the nucleus—a single, positively charged proton. Looking out into the vast emptiness of the atom, you see an electron zipping around you in a circular path. This is the $n^{	ext{th}}$ Bohr orbit. But how fast is this electron moving? Does it speed up or slow down as it moves to orbits further away? Let's find out!

### The Master Equation
Niels Bohr, in his brilliant model of the atom, gave us a way to calculate the exact velocity of an electron in any orbit. By balancing the electrostatic force of attraction with the required centripetal force, and applying his quantization of angular momentum ($mvr = \frac{nh}{2\pi}$), we arrive at a beautiful expression for the velocity $v_n$:

$$v_n = \frac{Ze^2}{2\varepsilon_0 nh}$$

I know this equation might look a bit intimidating at first glance, but let's take a breath and break it down.

### Decoding the Constants
Look closely at the terms in our master equation. For a given atom, like hydrogen, the atomic number $Z$ is a constant ($Z=1$). The charge of the electron $e$, the permittivity of free space $\varepsilon_0$, and Planck's constant $h$ are all universal constants. They never change!

So, if we group all these constants together, we can rewrite our equation as:

$$v_n = (	ext{Constant}) 	imes \frac{1}{n}$$

### The Final Verdict
This reveals a profound physical truth: the velocity of the electron $v_n$ is **inversely proportional** to the principal quantum number $n$.

$$v_n \propto \frac{1}{n}$$

What does this mean physically? As the electron jumps to higher orbits (larger $n$), it moves further away from the nucleus. The electrostatic pull from the nucleus weakens, so the electron doesn't need to travel as fast to maintain its orbit. It slows down! This elegant relationship perfectly matches option (b).

Always remember this inverse proportionality—it's a classic concept that frequently appears in JEE!

Animated Solution for Physics - Atoms and Nuclei: If an electron is moving in the $n$ th orbit of the hydrogen atom, then its velocity $v_{n}$ for the $n$ th orbit is given as

Select Answer:

Visualized Solution

Bohr's Model of Hydrogen Atom

Velocity of Electron in $n^{th}$ Orbit

Identifying Constants

Proportionality

Further Analysis

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Cosmic Dance of the Electron Imagine you are standing at the center of a hydrogen atom

The Master Equation Niels Bohr, in his brilliant model of the atom, gave us a way to calculate the exact velocity of an electron in any orbit

Decoding the Constants Look closely at the terms in our master equation

The Final Verdict

Similar Questions

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = \frac{1}{2} k r^{2}$ . If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as

In a hydrogen atom, electron makes a transition from $(n + 1)$ th level to the $n$ th level. If $n >> 1$ , the frequency of radiation emitted is proportional to

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \overset{˚}{A}$ ). The de-Broglie wavelength of this electron is

In a hydrogen like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$ . If $n >> 1$ , the frequency of radiation emitted is proportional to

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = U_{0} r^{4}$ . If Bohr's quantisation conditions are applied, radii of possible orbitals $r_{n}$ vary with $n^{1/ α}$ , where $α$ is .................

In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state $n$ is ........... .

Consider a hydrogen atom with its electron in the $n^{th}$ orbit. An electromagnetic radiation of wavelength $90 nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 eV$ , then the value of $n$ is ( $h c = 1242 eV nm$ )

In the Bohr model of the hydrogen atoms

Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an electron is moving in the nth orbit of the hydrogen atom, then its velocity vn​ for the nth orbit is given as