
In a hydrogen atom following Bohr's postulates, the product of linear momentum and angular momentum is proportional to \[{n^x}\] where n is the orbit number. Then x is
A. 0
B. 2
C. -2
D. 1
Answer
163.2k+ views
Hint: In order to solve this problem, we need to find the linear momentum and the angular momentum of the electron in the hydrogen atom as per Bohr’s postulates.
Formula used:
\[{v_n} = \left( {2.18 \times {{10}^6}m/s} \right)\left( {\dfrac{Z}{n}} \right)\]
where \[{v_n}\] is the speed of an electron in the nth orbit of an atom with atomic number Z.
\[L = \dfrac{{nh}}{{2\pi }}\]
where L is the angular momentum of an electron in the nth orbit of an atom.
Complete step by step solution:
According to Bohr’s model of atoms, the electron revolves around the nucleus of the atom with uniform speed. In a hydrogen atom, there is only one electron revolving around the nucleus under the influence of attractive force applied by the positive nucleus of the hydrogen.
The speed of the electron in nth orbit of atom is given as,
\[{v_n} = \left( {2.18 \times {{10}^6}\dfrac{Z}{n}} \right)m/s\]
Here, Z is the atomic number of the atom.
For hydrogen atoms, the atomic number is 1.
So, the speed of electron in nth orbit of hydrogen atom will be,
\[{v_n} = \dfrac{{\left( {2.18 \times {{10}^6}m/s} \right)}}{n}\]
So, the linear momentum of the electron will be,
\[P = m{v_n}\]
\[\Rightarrow P = m\left( {\dfrac{{2.18 \times {{10}^6}}}{n}} \right) \\ \]
The angular momentum of the electron as per Bohr’s postulates is the integral multiple of \[\dfrac{h}{{2\pi }}\]
For the electron in hydrogen atom in nth orbit, the angular momentum will be,
\[L = \dfrac{{nh}}{{2\pi }}\]
Then the product of the linear momentum of the electron in hydrogen atom and the angular momentum of the electron in the same energy level of hydrogen atom is,
\[LP = \left( {\dfrac{{nh}}{{2\pi }}} \right)\left( {\dfrac{{2.18 \times {{10}^6}m}}{n}} \right) \\ \]
\[\Rightarrow LP = {n^0}\left( {\dfrac{{2.18 \times {{10}^6}mh}}{{2\pi }}} \right) \\ \]
So, the product of the linear momentum of the electron in hydrogen atom and the angular momentum is proportional to \[{n^0}\], but in the question is given that the product of the linear momentum and the angular momentum of the electron in the hydrogen atom is \[{n^x}\]. On comparing both the terms, we get \[x = 0\]. Hence, the value of x is zero.
Therefore,the correct option is A.
Note: The angular momentum is always perpendicular to the linear momentum vector. Bohr's postulate is valid for hydrogen or hydrogen line atoms. The hydrogen- like atom is an atom with a single electron orbiting the nucleus.
Formula used:
\[{v_n} = \left( {2.18 \times {{10}^6}m/s} \right)\left( {\dfrac{Z}{n}} \right)\]
where \[{v_n}\] is the speed of an electron in the nth orbit of an atom with atomic number Z.
\[L = \dfrac{{nh}}{{2\pi }}\]
where L is the angular momentum of an electron in the nth orbit of an atom.
Complete step by step solution:
According to Bohr’s model of atoms, the electron revolves around the nucleus of the atom with uniform speed. In a hydrogen atom, there is only one electron revolving around the nucleus under the influence of attractive force applied by the positive nucleus of the hydrogen.
The speed of the electron in nth orbit of atom is given as,
\[{v_n} = \left( {2.18 \times {{10}^6}\dfrac{Z}{n}} \right)m/s\]
Here, Z is the atomic number of the atom.
For hydrogen atoms, the atomic number is 1.
So, the speed of electron in nth orbit of hydrogen atom will be,
\[{v_n} = \dfrac{{\left( {2.18 \times {{10}^6}m/s} \right)}}{n}\]
So, the linear momentum of the electron will be,
\[P = m{v_n}\]
\[\Rightarrow P = m\left( {\dfrac{{2.18 \times {{10}^6}}}{n}} \right) \\ \]
The angular momentum of the electron as per Bohr’s postulates is the integral multiple of \[\dfrac{h}{{2\pi }}\]
For the electron in hydrogen atom in nth orbit, the angular momentum will be,
\[L = \dfrac{{nh}}{{2\pi }}\]
Then the product of the linear momentum of the electron in hydrogen atom and the angular momentum of the electron in the same energy level of hydrogen atom is,
\[LP = \left( {\dfrac{{nh}}{{2\pi }}} \right)\left( {\dfrac{{2.18 \times {{10}^6}m}}{n}} \right) \\ \]
\[\Rightarrow LP = {n^0}\left( {\dfrac{{2.18 \times {{10}^6}mh}}{{2\pi }}} \right) \\ \]
So, the product of the linear momentum of the electron in hydrogen atom and the angular momentum is proportional to \[{n^0}\], but in the question is given that the product of the linear momentum and the angular momentum of the electron in the hydrogen atom is \[{n^x}\]. On comparing both the terms, we get \[x = 0\]. Hence, the value of x is zero.
Therefore,the correct option is A.
Note: The angular momentum is always perpendicular to the linear momentum vector. Bohr's postulate is valid for hydrogen or hydrogen line atoms. The hydrogen- like atom is an atom with a single electron orbiting the nucleus.
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?
