
The energy of an electron in an excited hydrogen atom is −3.4 eV. Then, according to Bohr's Theory, the angular momentum of this electron, in J s, is
A. \[2.11 \times {10^{ - 34}}\]
B. \[3 \times {10^{ - 34}}\]
C. \[1.055 \times {10^{ - 34}}\]
D. \[0.5 \times {10^{ - 34}}\]
Answer
163.5k+ views
Hint: An electron's angular momentum in the orbit is quantized. Find out the orbit number in which an electron is situated by putting the value of given energy in the formula \[E = - \dfrac{{13.6}}{{{n^2}}}eV\]. Then when we get value of $n$ and can easily find out the angular momentum by the formula \[mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}\].
Formula used:
Energy of the electron in nth orbit is,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
The formula of angular momentum ($L$) is,
\[L= {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Here, $h$ is the planck’s constant and $n$ is the orbit number.
Complete step by step solution:
Three postulates form the foundation of Bohr's model of the hydrogen atom:
(1) Around the nucleus, an electron moves in a circular orbit.;
(2) The quantization of an electron's orbital angular momentum; and
(3) The electrons in an atom move from a lower energy level to a higher energy level by gaining the necessary energy, and an electron moves from a higher energy level to a lower energy level by losing energy.
Let's discuss the second postulates in more detail. The second premise discusses stable orbits. Bohr claimed that electrons can rotate in specific discrete (discontinuous), non-radiating orbits known as stationary (permitted) orbits, for which the angular momentum of the rotating electron (L) is an integral multiple of $\dfrac{h}{{2\pi }}$.
As a result, the circling electron's angular momentum is an integral multiple of integer. An electron's angular momentum is equal to \[mvr\]. For every orbit that is stationary:
\[mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Where, n = a main quantum number with a positive integral value between 1 and n with a range of 1, 2, and 3. Planck's constant, h, is equal to \[6.63{\text{ }} \times {\text{ }}{10^{ - 34}}{\text{ }}Js\].
The energy in any orbit for a hydrogen atom can be defined as,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
Here energy is given -3.4 eV. So, we can write
\[E = \dfrac{{ - 13.6}}{{{n^2}}}eV \\
\Rightarrow - 3.4 = - \dfrac{{13.6}}{{{n^2}}} \\ \]
$\Rightarrow {n^2} = \dfrac{{ - 13.6}}{{ - 3.4}} = 4 \\
\Rightarrow n = 2 \\ $
Now we have to find out the angular momentum of Bohr’s second’s orbit. So, we already see a formula for finding angular momentum. If we put the given values then we get.
${\text{Angular momentum = }}\dfrac{{{\text{nh}}}}{{2\pi }} \\
\Rightarrow \text{Angular momentum} = \dfrac{{2 \times 6.63 \times {{10}^{ - 34}}}}{{2 \times 3.14}} \\
\therefore \text{Angular momentum}= 2 \times {10^{ - 34}}kg - {m^2}/\sec $
Hence, option A is the correct option.
Notes The primary distinction between angular momentum and linear momentum is that linear momentum is a property of moving objects relative to a reference point (i.e., any object changing its position with respect to the reference point), angular momentum is a property of moving objects relative to a reference point that also involves changing their position's direction (i.e. they are not moving in a straight line).
Formula used:
Energy of the electron in nth orbit is,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
The formula of angular momentum ($L$) is,
\[L= {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Here, $h$ is the planck’s constant and $n$ is the orbit number.
Complete step by step solution:
Three postulates form the foundation of Bohr's model of the hydrogen atom:
(1) Around the nucleus, an electron moves in a circular orbit.;
(2) The quantization of an electron's orbital angular momentum; and
(3) The electrons in an atom move from a lower energy level to a higher energy level by gaining the necessary energy, and an electron moves from a higher energy level to a lower energy level by losing energy.
Let's discuss the second postulates in more detail. The second premise discusses stable orbits. Bohr claimed that electrons can rotate in specific discrete (discontinuous), non-radiating orbits known as stationary (permitted) orbits, for which the angular momentum of the rotating electron (L) is an integral multiple of $\dfrac{h}{{2\pi }}$.
As a result, the circling electron's angular momentum is an integral multiple of integer. An electron's angular momentum is equal to \[mvr\]. For every orbit that is stationary:
\[mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Where, n = a main quantum number with a positive integral value between 1 and n with a range of 1, 2, and 3. Planck's constant, h, is equal to \[6.63{\text{ }} \times {\text{ }}{10^{ - 34}}{\text{ }}Js\].
The energy in any orbit for a hydrogen atom can be defined as,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
Here energy is given -3.4 eV. So, we can write
\[E = \dfrac{{ - 13.6}}{{{n^2}}}eV \\
\Rightarrow - 3.4 = - \dfrac{{13.6}}{{{n^2}}} \\ \]
$\Rightarrow {n^2} = \dfrac{{ - 13.6}}{{ - 3.4}} = 4 \\
\Rightarrow n = 2 \\ $
Now we have to find out the angular momentum of Bohr’s second’s orbit. So, we already see a formula for finding angular momentum. If we put the given values then we get.
${\text{Angular momentum = }}\dfrac{{{\text{nh}}}}{{2\pi }} \\
\Rightarrow \text{Angular momentum} = \dfrac{{2 \times 6.63 \times {{10}^{ - 34}}}}{{2 \times 3.14}} \\
\therefore \text{Angular momentum}= 2 \times {10^{ - 34}}kg - {m^2}/\sec $
Hence, option A is the correct option.
Notes The primary distinction between angular momentum and linear momentum is that linear momentum is a property of moving objects relative to a reference point (i.e., any object changing its position with respect to the reference point), angular momentum is a property of moving objects relative to a reference point that also involves changing their position's direction (i.e. they are not moving in a straight line).
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
