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# Bohr Model Of The Hydrogen Atom - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Radius of orbit and velocity of electron is considered one the most difficult concept.

• Bohr's Model of hydrogen atom is considered one of the most asked concept.

• 51 Questions around this concept.

## Solve by difficulty

The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is:

Suppose an electron is attracted towards the origin by a force k / r where k is constant and r is the distance of the electron from the origin. By applying the Bohr model to this system, the radius of the nth orbital of the electron is found to be rand the kinetic energy of the electron to be Tn. Then which of the following is true?

When the electron in a hydrogen atom is excited from the $4^{\text {th }}$ stationary orbit to the $5^{\text {th }}$ stationary orbit, the change in the angular momentum of the electron is: (Planck's constant, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js} )$

Suppose an electron is attracted towards the origin by a force $\frac{\mathrm{k}}{\mathrm{r}}$, where $\mathrm{k}$ is a constant and $\mathrm{r}$ is the distance of the electron from the origin. By applying the Bohr model to this system, the radius of the nth orbital of the electron is found to be $r_n$ and the kinetic energy of the electron to be $T_n$. Then, which of the following is true?

If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze, then from Bohr's first postulate, the kinetic energy of the electron in C.G.S. system is equal to

Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Broglie wavelength $\lambda$ of that electron as

In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. It $a_{0}$ is the radius of the ground state orbit, m is the mass, $e$ is the charge on the electron, and $\varepsilon _{0}$ is the vacuum permittivity, the speed of the electron is

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When an electron in a hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of the electron is (Planck’s constant: $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$)?

Orbital acceleration of electrons is

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According to Bohr’s theory, the time average magnetic field at the center (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n=principal quantum number)

## Concepts Covered - 2

Bohr's Model of hydrogen atom

Bohr's Model of hydrogen atom:

Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge  $Z_e$ (called hydrogen-like atom)
Bohr's model is based on the following postulates-

(1). Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom

For electrons revolving in a stable orbit, the necessary centripetal force is provided by the coulomb's force

$\frac{mv_{n}^{2}}{r_{n}}=\frac{kze^{2}}{r_{n}^{2}}$
(2). Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2\pi}$ where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is $L=m v_{n} r_{n}=\frac{n h}{2 \pi} ; n=1,2,3...... \infty$
(3). Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by $h\nu = E_i - E_f$

Ei is the energy of the initial state and Ef is the energy of the final state. Also, Ei > Ef.

vn- speed of an electron in the  nth orbit

Radius of orbit and velocity of electron

Radius of orbit and velocity of the electron

Radius of the orbit: For an electron around a stationary nucleus the electrostatic force of attraction provides the necessary centripetal force.

$\begin{array}{l}{\text { ie. } \frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e) e}{r^{2}}=\frac{m v^{2}}{r} \quad \cdots \text { (i) }} \\ \\ {\text { also } m v r=\frac{n h}{2 \pi}}\end{array}$

From equations (i) and (ii) radius of r orbit

$\begin{array}{l}{r_{n}=\frac{n^{2} h^{2}}{4 \pi^{2} k Z m e^{2}}=\frac{n^{2} h^{2} \varepsilon_{0}}{ \pi m Z e^{2}}=0.53 \frac{n^{2}}{Z} \AA \quad\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)} \\ \\ {\Rightarrow r_{n} \propto \frac{n^{2}}{Z}}\end{array}\\ \implies r_n=0.53 \frac{n^2}{Z} \AA$

Speed of electron:

From the above relations, the speed of electrons in $n^{th}$ orbit can be calculated as
$v_{n}=\frac{2 \pi k Z e^{2}}{n h}=\frac{Z e^{2}}{2 \varepsilon_{0} n h}=\left(\frac{c}{137}\right) \frac{Z}{n}=2.2 \times 10^{6} \frac{Z}{n} m / \text {sec}$

where $\left(c=\text { speed of light }=3 \times 10^{8} \mathrm{m} / \mathrm{s}\right)$

## Study it with Videos

Bohr's Model of hydrogen atom
Radius of orbit and velocity of electron

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