
Electrons in a certain energy level \[n = {n_1}\] , can emit 3 spectral lines. When they are at another energy level, \[n = {n_2}\]. They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio of
A. 4:3
B. 3:4
C. 2:1
D. 1:2
Answer
134.4k+ views
Hint:The electron jumps to the lower energy level by radiating out energy in the form of spectral lines. The number of spectral lines is proportional to the possible number of transitions made by the electron to reach the ground state.
Formula used:
Using Bohr’s postulate the speed of electron in nth state is given as,
\[{v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}\]
The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
\[N = \dfrac{{n\left( {n - 1} \right)}}{2}\]
Complete step by step solution:
When an electron is in nth state and jumps to the ground state then it releases energy in the form of radiation and emits the spectral lines. The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
\[N = \dfrac{{n\left( {n - 1} \right)}}{2}\]
It is given that when the electron is in the state \[n = {n_1}\] then it emits a total of 3 spectral lines.
Putting in the expression for the number of spectral lines emitted, we get
\[3 = \dfrac{{{n_1}\left( {{n_1} - 1} \right)}}{2}\]
On solving the equation, we get
\[n_1^2 - {n_1} - 6 = 0\]
\[\Rightarrow n_1^2 - 3{n_1} + 2{n_1} - 6 = 0\]
\[\Rightarrow \left( {{n_1} - 3} \right)\left( {{n_1} + 2} \right) = 0\]
As the state of the electron is a positive whole number, so the value of \[{n_1}\] is 3.
Similarly for the state \[{n_2}\] the number of spectral lines emitted is 6. Putting in the expression for the number of spectral lines emitted, we get
\[6 = \dfrac{{{n_2}\left( {{n_2} - 1} \right)}}{2}\]
On solving the equation, we get
\[n_2^2 - {n_2} - 12 = 0\]
\[\Rightarrow n_2^2 - 4{n_2} + 3{n_2} - 12 = 0\]
\[\Rightarrow \left( {{n_2} - 4} \right)\left( {{n_1} + 3} \right) = 0\]
Hence, the value of \[{n_2}\] is 4.
Using Bohr’s postulate the speed of electron in nth state is given as,
\[{v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}\]
So, the ratio of the speeds of the electron in both the state is,
\[\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\dfrac{{2\pi KZ{e^2}}}{{{n_1}h}}}}{{\dfrac{{2\pi KZ{e^2}}}{{{n_2}h}}}} \\ \]
\[\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{n_2}}}{{{n_1}}} \\ \]
\[\therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{4}{3}\]
Hence, the ratio of the speed of the electron in both the states is 4:3.
Therefore, the correct option is A.
Note: We must be careful while choosing the solution of the quadratic equation. As the principal quantum number is a positive whole number, we need to choose only the positive solution of the quadratic equation.
Formula used:
Using Bohr’s postulate the speed of electron in nth state is given as,
\[{v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}\]
The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
\[N = \dfrac{{n\left( {n - 1} \right)}}{2}\]
Complete step by step solution:
When an electron is in nth state and jumps to the ground state then it releases energy in the form of radiation and emits the spectral lines. The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
\[N = \dfrac{{n\left( {n - 1} \right)}}{2}\]
It is given that when the electron is in the state \[n = {n_1}\] then it emits a total of 3 spectral lines.
Putting in the expression for the number of spectral lines emitted, we get
\[3 = \dfrac{{{n_1}\left( {{n_1} - 1} \right)}}{2}\]
On solving the equation, we get
\[n_1^2 - {n_1} - 6 = 0\]
\[\Rightarrow n_1^2 - 3{n_1} + 2{n_1} - 6 = 0\]
\[\Rightarrow \left( {{n_1} - 3} \right)\left( {{n_1} + 2} \right) = 0\]
As the state of the electron is a positive whole number, so the value of \[{n_1}\] is 3.
Similarly for the state \[{n_2}\] the number of spectral lines emitted is 6. Putting in the expression for the number of spectral lines emitted, we get
\[6 = \dfrac{{{n_2}\left( {{n_2} - 1} \right)}}{2}\]
On solving the equation, we get
\[n_2^2 - {n_2} - 12 = 0\]
\[\Rightarrow n_2^2 - 4{n_2} + 3{n_2} - 12 = 0\]
\[\Rightarrow \left( {{n_2} - 4} \right)\left( {{n_1} + 3} \right) = 0\]
Hence, the value of \[{n_2}\] is 4.
Using Bohr’s postulate the speed of electron in nth state is given as,
\[{v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}\]
So, the ratio of the speeds of the electron in both the state is,
\[\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\dfrac{{2\pi KZ{e^2}}}{{{n_1}h}}}}{{\dfrac{{2\pi KZ{e^2}}}{{{n_2}h}}}} \\ \]
\[\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{n_2}}}{{{n_1}}} \\ \]
\[\therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{4}{3}\]
Hence, the ratio of the speed of the electron in both the states is 4:3.
Therefore, the correct option is A.
Note: We must be careful while choosing the solution of the quadratic equation. As the principal quantum number is a positive whole number, we need to choose only the positive solution of the quadratic equation.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

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